Math in Focus

Practice the problems of Math in Focus Grade 8 Workbook Answer Key Chapter 5 Lesson 5.5 Inconsistent and Dependent Systems of Linear Equations to score better marks in the exam.

Math in Focus Grade 7 Course 3 A Chapter 5 Lesson 5.5 Answer Key Inconsistent and Dependent Systems of Linear Equations

Math in Focus Grade 8 Chapter 5 Lesson 5.5 Guided Practice Answer Key

Identify whether each system of linear equations is inconsistent or has a unique solution. Justify your reasoning.

Question 1.
11x + y = 2 — Equation 1
22x + 2y = 3 —Equation 2
Rewrite each linear equations in slope-intercept form y = mx + b.
Equation 1
11x + y = 2
= Subtract 11x from both sides.
y = Simplify.
Equation 2
22x + 2y = 3
= Subtract 22x from both sides.
= Simplify.
= Divide both sides by 2.
y = Simplify.
The slope of the graph of the Equation 1 is and the y-intercept is .
The slope of the graph of the Equation 2 is and the y-intercept is .
The graphs of the linear equations have the same and different . The system of linear equations is , because the system of linear equations has no solution.
Answer:
11x + y = 2 Equation 1 We are given the system of equations:
22x + 2y = 3 Equation 2
11x + y = 2 Rewrite Equation 1 in slope-intercept form y = mx + b:
11x + y — 11x = 2 — 11x Subtract 11x from both sides:
y = —11x + 2 Simplify:
22x + 2y = 3 Rewrite Equation 2 in slope-intercept form y = mx + b
22x + 2y — 22x = 3 — 22x Subtract 22x from both sides
2y = —22x + 3 Simplify:
\(\frac{2 y}{2}\) = \(\frac{-22 x+3}{2}\) Divide both sides by 2:
y = -11x + \(\frac{3}{2}\) Simplify:
The slope of the graph of Equation 1 is -11 and the y-intercept is 2.
The slope of the graph of Equation 2 is -11 and the y-intercept is \(\frac{3}{2}\).
The graphs of the linear equations have the same slope and different y-intercept. The system of linear equations is inconsistent because the system of linear equations has no solution.
Inconsistent

Question 2.
13x + y = 23
26x + 2y = 21
Answer:
13x + y = 23 Equation 1 We are given the system of equations:
26x + 2y = 21 Equation 2
13x + y = 23 Rewrite Equation 1 in slope-intercept form y = mx + b:
13x + y — 13x = 23 — 13x Subtract 13x from both sides
y = —13x -i- 23 Simplify:
26x + 2y = 21 Rewrite Equation 2 in slope-intercept form y = mx + b:
26x + 2y — 26x = 21 — 26x Subtract 26x from both sides:
2y = -26x + 21 Simplify:
\(\frac{2 y}{2}\) = \(\frac{-26 x+21}{2}\) Divide both sides by 2:
y = —13x + \(\frac{21}{2}\) Simplify:
The slope of the graph of Equation 1 is -13 and the gj-intercept is 23.
The slope of the graph of Equation 2 is -13 and the y-intercept is \(\frac{21}{2}\).
The graphs of the linear equations have the same slope and different y-intercept The system of linear equations is inconsistent because the system of linear equations has no solution.
Inconsistent

Question 3.
\(\frac{1}{2}\)x + y = 4
3x + 8y = 16
Answer:
\(\frac{1}{4}\)x + y = 4 Equation 1
3x + 8y = 16 Equation 2
We are given the system of equations:
\(\frac{1}{2}\)x + y = 4 Rewrite Equation 1 in slope-intercept form y = mx + b:
\(\frac{1}{2}\)x + y – \(\frac{1}{2}\)x = 4 – \(\frac{1}{2}\)x Subtract 11x from both sides:
y = – \(\frac{1}{2}\)x + 4 Simplify:
3x + 8 = 16 Rewrite Equation 2 in slope-intercept form y = mx + b:
3x + 8y — 3x = 16 — 3x Subtract 3x from both sides:
8y = —3x + 16 Simplify:
\(\frac{8 y}{8}\) = \(\frac{-3 x+16}{8}\) Divide both sides by 8:
y = \(-\frac{3}{8}\)x + 2 simplify
The slope of the graph of Equation 1 is – \(\frac{1}{2}\) and the y-intercept is 4.
The slope of the graph of Equation 2 is – \(\frac{3}{8}\) and the y-intercept is 2.
– \(\frac{1}{2}\)x + 4 = — \(\frac{3}{8}\) + 2 The graphs of the linear equations have different slopes, therefore tne system of linear equations has a unique solution.
8 ∙ (-\(\frac{1}{2}\)x + 4) = 8 ∙ (-\(\frac{3}{8}\)x + 2
—4x + 32 = -3x + 16
-4x + 32 + 4x = -3x + 16 + 4x
32 = x + 16
32 — 16 = x + 16 — 16
x = 16
y = –\(\frac{3}{8}\)(16) + 2
= -6 + 2
= -4
Unique solution (x = 16; y = -4)

Question 4.
\(\frac{1}{3}\)x + y = 7
x + 3y = 6
Answer:
\(\frac{1}{3}\)x + y = 7 Equation 1
x + 3y = 6 Equation 2
We are given the system of equations:
\(\frac{1}{3}\)x + y = 7 Rewrite Equation lin slope-intercept form y = mx + b:
\(\frac{1}{3}\)x + y – \(\frac{1}{3}\)x = 7 – \(\frac{1}{3}\)x Subtract \(\frac{1}{3}\)x from both sides:
y = –\(\frac{1}{3}\)x + 7 Simplify:
x + 3y = 6 Rewrite Equation 2 in slope-intercept form y = mx + b:
x + 3y — x = 6 — x Subtract x from both sides:
3y = —x + 6 Simplify:
\(\frac{3 y}{3}\) = \(\frac{-x+6}{3}\) Divide both sides by 3:
y = \(\frac{1}{3}\)x + 2 Simplify:
The slope of the graph of Equation 1 is —\(\frac{1}{3}\) and the y-intercept is 7
The slope of the graph of Equation 2 is –\(\frac{1}{3}\) and the y-irtercept is 2.
The graphs of the linear equations have the same slope and different y-intercept The system of linear equations is inconsistent because the system of linear equations has no solution.
Inconsistent

Identify whether the system of linear equations is inconsistent or has a unique solution. Justify your reasoning.

Question 5.
5x + 2y = 4
20x + 8y = 30
Answer:
5x + 2y = 4 Equation 1
20x + 8y = 30 Equation 2
We are given the system of equations:
5x + 2 = Rewrite Equation 1 in slope-intercept form y = mx + b:
5x + 2y — 5x = 4 — 5x Subtract 5x from both sides and simplify:
\(\frac{2 y}{2}\) = \(\frac{-5 x+4}{2}\)
y = \(-\frac{5}{2}\)x + 2 Divide both sides by 2 and simplify:
20x + 8y = 30 Rewrite Equation 2 in slope-intercept form y = mx + b:
20x + 8y – 20x = 30 — 20x Subtract 20x from both sides and simplify:
8y = -20x + 30
\(\frac{8 y}{8}\) = \(\frac{-20 x+30}{8}\) Divide both sides by 8 and simplify:
y = –\(\frac{5}{2}\)x + \(\frac{15}{4}\)
The slope of the graph of Equation 1 is –\(\frac{5}{2}\) and the y-intercept is 2.
The slope of the graph of Equation 2 is —\(\frac{5}{2}\) and the y-intercept is \(\frac{15}{4}\).
The graphs of the linear equations have the same slope and different y-intercept The system of linear equations is inconsistent because the system of linear equations has no solution.
Inconsistent

Question 6.
7x + y = 14
10x + 4y = 32
Answer:
7x + y = 14 Equation 1 We are given the system of equations
10x + 4y = 32 Equation 2
7x + 4y = 14 Rewrite Equation 1 in slope-Intercept form y = mx + b
7x + y — 7x = 14 — 7x subtract 7x from both sides and simplify
y = —7x + 14 simplify
10x + 4y = 32 Rewrite Equation 2 in slope-intercept form y = mx + b
10x + 4y – 10x = 32 — 10x Subtract 10x from both sides and simplify:
4y = -10x + 32 simplify
\(\frac{4 y}{2}\) = \(\frac{-10 x+32}{4}\) Divide both sides by 4
y = –\(\frac{5}{2}\)x + 8 Simplify:
The slope of the graph of Equation 1 is -7 and the y-intercept is 14.
The slope of the graph of Equation 2 is –\(\frac{5}{2}\) and the y-intercept is 8.

—7x + 14 = —\(\frac{5}{2}\)x + 8 The graphs of the linear equations have different slopes, therefore the system of linear equations has an unique solution.
2 ∙ (-7x + 14) = 2(-\(\frac{5}{2}\)x + 8)
—14x + 28 = —5x + 16
—14x + 28 + 14x = —5x + 16 + 14x
28 = 9x + 16
28 — 16 = 9x + 16 — 16
12 = 9x
\(\frac{12}{9}\) = \(\frac{9 x}{9}\)
x = \(\frac{4}{3}\)
y = –\(\frac{5}{2}\) ∙ \(\frac{4}{3}\) + 8
= –\(\frac{10}{3}\) + 8
= \(\frac{14}{3}\)
Unique solution (x = \(\frac{4}{3}\); y = \(\frac{14}{3}\))

Question 7.
12x + 4y = 16
9x + 3y = 12
Answer:
12x + 4y = 16 Equation 1
9x + 3y = 12 Equation 2
we are given the system of equations:
12x + 4y = 16 Rewrite Equation 1 in slope-intercept form y = mx + b:
12x + 4y — 12x = 16 — 2x Subtract 12x from both sides and simplify:
4y = —12x + 16
\(\frac{4 y}{4}\) = \(\frac{-12 x+16}{4}\) Divide both sides by 4 and simplify:
y = —3x + 4
9x + 3y = 12 Rewrite Equation 2 in slope-intercept form y = mx + b:
9x + 3y — 9x = 12 — 9x Subtract 9x from both sides and simplify:
3y = —9x + 12
\(\frac{3 y}{4}\) = \(\frac{-9 x+12}{3}\) Divide both sides by 3 and simplify:
y = -3x + 4
Since Equation 1 and Equation 2 are equivalent, they have an infinite number of solutions. Therefore system of linear equations is dependent.
Dependent

Math in Focus Course 3A Practice 5.5 Answer Key

Graph each system of linear equations using a graphing calculator. State whether each system of equations is inconsistent or has a unique solution.

Question 1.
3x + y = 4
6x + 2y = 14
Answer:
3x + y = 4 Equation 1 We are given the system of equations:
6x + 2y = 14 Equation 2
3x + = 4 Rewrite Equation 1 in slope-intercept form y = mx + b:
3x + y — 3x = 4 — 3x Subtract 3x from both sides:
y = —3x + 4 Simplify:
6x + 2y = 14 Rewrite Equation 2 in slope-intercept form y = mx + b:
6x + 2y — 6x = 14 — 6x Subtract 6x from both sides:
2y = —6x + 14 Simplify:
\(\frac{2 y}{2}\) = \(\frac{-6 x+14}{2}\) Divide both sides by 2:
y = -3x + 7 Simplify:

We graph the two equations using a graphing calculator:
The graphs of the linear equations are parallel tines, so there is no point of intersection. Since there are no points that lie on both graphs, the system of linear equations is inconsistent
Inconsistent

Question 2.
10x + 5y = 15
x + y = 3
Answer:
10x + 5y = 15 Equation 1
x + y = 3 Equation 2
We are given the system of equations:
10x + 5y = 15 Rewrite Equation 1 in slope-intercept form y = mx + b:
10x + 5y — 10x = 15 — 10x Subtract 10x from both sides and simplify:
5y = 15 — 10x
\(\frac{5 y}{5}\) = \(\frac{-10 x+15}{5}\) Divide by 5 and simplify:
y = —2x + 3 Rewrite Equation 2 in slope-intercept form y = mx + b:
x + y — x = 3 — x Subtract x from both sides and simplify:

We graph the two equations:
x = 0
The graphs intersect in the point (0, 3). The system of linear equations has an unique solution:
y = 3
Unique solution (x = 0, y = 3)

Question 3.
x – 2y = 14
4y = 5 + 2x
Answer:
x — 2y = 14 Equation 1
4y = 5 + 2x Equation 2
We are given the system of equations:
x — 2y = 11 Rewrite Equation 1 in slope-intercept form y = mx + b:
x — 2y — x = 14 — x Subtract x from both sides and simplify:
-2y = 14 – x
\(\frac{-2 y}{-2}\) = \(\frac{14-x}{-2}\) Divide by -2 and simplify:
y = \(\frac{1}{2}\)x – 7
Divide by -2 and simplify:
4y = 5 + 2x Rewrite Equation 2 in slope-intercept form y = mx + b:
\(\frac{4 y}{4}\) = \(\frac{2 x+5}{4}\) Divide both sides by 4 and simplify:
y = \(\frac{1}{2}\)x + \(\frac{5}{4}\)

Graph both equations using a graphing calculator:
The graphs of the linear equations are parallel therefore there is no point to lie on both graphs, so the system has no solution, it is inconsistent.
Inconsistent

State whether each system of linear equations is inconsistent, dependent, or has a unique solution. Justify each answer. Solve each system of linear equations if possible.

Question 4.
x + y = 3
8x + 8y = 32
Answer:
x + y = 3 Equation 1
8x + 8y = 32 Equation 2
We are given the system of equations:
x + y = 3 Rewrite Equation 1 in slope-intercept form y = mx + b:
x + y — x = 3 — x Subtract x from both sides and simplify:
y = -x + 3
8x + 8y = 32 Rewrite Equation 2 in slope-intercept form y = mx + b:
8x + 8y — 8x = 32 — 8x Subtract 8x from both sides and simplify:
\(\frac{8 y}{8}\) = \(\frac{-8 x+32}{8}\) Divide both sides by 8 and simplify:
y = -x + 4
The linear equations have the same slope and different y-intercept The system of linear equations is inconsistent because the system of linear equations has no solution.
Inconsistent

Question 5.
6x + 2y = 12
3x + y = 21
Answer:
6x + 2y = 12 Equation 1
3x + y = 21 Equation 2
We are given the system of equations:
6x + 2y = 12 Rewrite Equation 1 in slope-intercept form y = mx + b:
6x + 2y — 6x = 12 — 6x Subtract 6x from both sides and simplify:
2y = —6x + 12
\(\frac{2 y}{2}\) = \(\frac{-6 x+12}{2}\) Divide both sides by 2 and simplify:
y = —3x + 6
3x + y = 21 Rewrite Equation 2 in slope-intercept form y = mx + b
3x + y – 3x = 21 – 3x Subtract 3x from both sides and simplify
y = -3x + 21
The linear equations have the same slope and different y-Intercept. The system of linear equations is Inconsistent because the system of linear equations has no solution.
Inconsistent

Question 6.
3x – 6y = 12
x – 2y = 4
Answer:
3x — 6y = 12 Equation 1
x — 2y = 4 Equation 2
We are given the system of equations:
3x – 6y = 12 Rewrite Equation 1 in slope-intercept form y = mx + b:
3x — 6y — 3x = 12 — 3x Subtract 3x from both sides and simplify:
—6y = —3x + 12
\(\frac{-6 y}{-6}\) = \(\frac{-3 x+12}{-6}\) Divide both sides by -6 and simplify:
y = \(\frac{1}{2}\)x – 2
x – 2y = 4 Rewrite Equation 2 in slope-intercept form y = mx + b:
x — 2y — x = 4 — x Subtract x from both sides and simplify:
—2y = —x + 4
\(\frac{-2 y}{-2}\) = \(\frac{-x+4}{-2}\) Divide both sides by -2 and simplify:
y = \(\frac{1}{2}\)x – 2
Since Equation 1 and Equation 2 are equivalent, they have an infinite number of solutions.
Therefore system of linear equations is dependent
Dependent

Question 7.
-15x + 3y = 3
-5x + y = 21
Answer:
—15x + 3y = 3 Equation 1
—5x + y = 21 Equation 2
We are given the system of equations:
-15x + 3y = 3 Rewrite Equation 1 in slope-intercept form y = mx + b:
—15x + 3y — 15x = 3 + 15x Add 15x to both sides and simplify
3y = 15x + 3
\(\frac{3 y}{3}\) = \(\frac{15 x+3}{3}\) Divide both sides by 3 and simplify:
y = 5x + 1
-5x + y = 21 Rewrite Equation 2 in slope-intercept form y = mx + b:
—5x + y + 5x = 21 + 5x Add 5x to both sides and simplify:
y = 5x + 21
The linear equations have the same slope and different y-intercept. The system of linear equations is inconsistent because the system of linear equations has no solution.
Inconsistent

Question 8.
2x + y = 8
4x – 2y = 24
Answer:
2x + y = 8 Equat ion 1
4x – 2y = 24 Equation 2
We are given the system of equations:
2x + y = 8 Rewrite Equation 1 in slope-intercept form y = mx + b:
2x + y — 2x = 8 — 2x Subtract 2x from both sides and simplify:
y = —2x + 8
4x — 2y = 24 Rewrite Equation 2 in slope-intercept form y = mx + b:
4x — 2y — 4x = 24 — 4x Subtract 4x from both sides and simplify:
—2y = —4x + 24
\(\frac{-2 y}{-2}\) = \(\frac{-4 x+24}{-2}\) Divide both sides by -2 and simplify:
y = 2x — 12
The slope of the graph of Equation 1 is -2 and the y-intercept is 8.
The slope of the graph of Equation 2 is 2 and the y-intercept is —12.
—2x + 8 = 2x — 12 The graphs of the linear equations have different slopes, therefore the system of linear equations has an unique solution.
—2x + 8 + 2x = 2x – 12 + 2x
8 = 4x — 12
8 + 12 = 4x — 12 + 12
4x = 20
x = \(\frac{20}{4}\)
x = 5
y = -2(5) + 8
= —10+8
= -2
Unique solution (x = 5; y = -2)

Question 9.
8x + 7y = 9
40x + 35y = 45
Answer:
8x + 7y = 9 Equation 1
40x + 35y = 45 Equation 2
We are given the system of equations:
8x + 7y = 9 Rewrite Equation 1 in slope-intercept form y = mx + b:
8x + 7y – 8x = 9 — 8x Subtract 8x from both sides and simplify:
7y = -8x + 9
\(\frac{7 y}{7}\) = \(\frac{-8 x+9}{7}\)
y = –\(\frac{8}{7}\)x + \(\frac{9}{7}\) Divide both sides by 7 and simplify:
40x + 35y = 45 Rewrite Equation 2 in slope-intercept form y = mx + b:
40x + 35y — 40x = 45 — 40x Subtract 40 from both sides and simplify:
35y = —40x + 45
\(\frac{35 y}{35}\) = \(\frac{-40 x+45}{35}\) Divide both sides by 35 and simpbfy:
y = –\(\frac{8}{7}\)x + \(\frac{9}{7}\)x
Since Equation 1 and Equation 2 are equivalent, they have an infinite number of solutions. Therefore system of linear equations is dependent
Dependent

Question 10.
x + 3y = 9
2x + 6y = 5
Answer:
x + 3y = 9 Equation 1
2x + 6y = 5 Equation 2
We are given the system of equations:
x + 3y = 9 Rewrite Equation 1 in slope-intercept form y = mx + b:
x + 3y — x = 9 — x Subtract x from both sides and simplify:
\(\frac{3 y}{3}\) = \(\frac{-x+9}{3}\) Divide both sides by 3 and simplify:
y = \(-\frac{1}{3} x\) + 3
2x + 6y = 6
Rewrite Equation 2 in slope-intercept form y = mx + b:
2x + 6y — 2x = 5 — 2x Subtract 2x from both sides and simplify:
6y = —2x + 5
\(\frac{6 y}{6}\) = \(\frac{-2 x+5}{6}\) Divide both sides by 6 and simplify:
y = –\(\frac{1}{3} x\) + \(\frac{5}{6}\)
The linear equations have the same slope and different y-intercept. The system of linear equations is inconsistent because the system of linear equations has no solution.
Inconsistent

Question 11.
9x + 21y = 27
6x + 14y = 18
Answer:
9x + 21y = 27 Equation 1
6x + 14y = 18 Equation 2
We are given the system of equations:
9x + 21y = 27 Rewrite Equation 1 in slope-intercept form y = mx + b:
9x + 21y – 9x = 27 — 9x Subtract 9x from both sides and simplify:
21y = —9x + 27
\(\frac{21 y}{21}\) = \(\frac{-9 x+27}{21}\) Divide both sides by 21 and simplify:
y = \(-\frac{3}{7}\)x + \(\frac{9}{7}\)
6x + 14y = 18 Rewrite Equation 2 in slope-intercept form y = mx + b
6x + 14y — 6x = 18 — 6x Subtract 6x from both sides and simplify:
14y = —6x + 18
\(\frac{14 y}{14}\) = \(\frac{-6 x+18}{14}\) Divide both sides by 14 and simplify:
y = \(-\frac{3}{7}\) + \(\frac{9}{7}\)
Since Equation 1 and Equation 2 are equivalent, they have an infinite number of solutions.
Therefore system of linear equations is dependent
Dependent

Question 12.
5x + 4y = 6
15x + 12y = 18
Answer:
5x + 4y = 6 Equation 1
15x + 12y = 18 Equation 2
We are given the system of equations:
5x + 4y = 6 Rewrite Equation 1 in slope-intercept form y = mx + b:
5x + 4y — 5x = 6 — 5x Subtract 4x from both sides and simplify:
4y = -5x + 6
\(\frac{4 y}{4}\) = \(\frac{-5 x+6}{4}\) Divide both sides by 5 and simplify:
y = \(-\frac{5}{4}\)x + \(\frac{3}{2}\)
15x + 12y = 18 Rewrite Equation 2 in slope-intercept form y = mx + b:
15x + 12y — 15x = 18 — 15x Subtract 15x from both sides and simplify:
12y = -15x + 18
\(\frac{12 y}{12}\) = \(\frac{-15 x+18}{12}\) Divide both sides by 12 and simplify:
y = \(-\frac{5}{4}\)x + \(\frac{3}{2}\)
Since Equation 1 and Equation 2 are equivalent, they have an infinite number of solutions.
Therefore system of linear equations is dependent
Dependent

Solve. Show your work.

Question 13.
David says he bought 9 apples and 6 apricots for $8.50 yesterday and bought 3 apples and 2 apricots for $7.40 today.

a) Write a system of equations to find the cost of an apple and an apricot.
Answer:
9x + 6y = 8.50 Equation 1
3x + 2y = 7.40 Equation 2
a) Lets note:
x = the cost of an apple
y = the cost of an apricot
We can write the system of equations:

b) State with reasons whether the system of equations has a unique solution, is inconsistent, or is dependent.
Answer:
9x + 6y = 8.50 b) Rewrite Equation 1 in slope-intercept form y = mx + b
9x + 6y — 9x = 8.50 – 9x Subtract 9x from both sides and simplify:
6y = -9x + 8.50
\(\frac{6 y}{6}\) = \(\frac{-9 x+8.50}{6}\) Divide both sides by 6 and simplify:
y = \(-\frac{3}{2}\)x + \(\frac{4.25}{3}\)
3x + 2y = 7.40 Rewrite Equation 2 in slope-intercept form y = mx + b:
3x + 2y — 3x = 7.40 — 3x Subtract 3x from both sides and simplify:
2y = 7.40 — 3x
\(\frac{2 y}{2}\) = \(\frac{-3 x+7.40}{2}\) Divide both sides by 2 and simplify:
y = –\(\frac{3}{2}\) + 3.70
The linear equations have the same slope and different y-intercept. The system of linear equations is inconsistent because the system of linear equations has no solution.

c) What does this tell you about the cost of apples and apricots on those two days?
Answer:
The cost of at least one type of fruits is different in the two days.

a) 9x + 6y = 8.50
3x + 2y = 7.40
b) Inconsistent system
c) The unit price changed

Question 14.
Ms. Cohen gave a riddle: A string is 2 meters longer than a rod. Half of the rod is 1 meter shorter than half of the string. Is this true or false?
a) Write a system of equations to find the length of the string and the length of the rod.
Answer:
s = r + 2 Equation 1
\(\frac{1}{2}\)r = \(\frac{1}{2}\)s – 1 Equation 2
(a) Let’s note:
s=the length of a string
r=the length of a rod
We can write the system of equations:

b) State with reasons whether the system of equations has a unique solution, is inconsistent, or is dependent.
Answer:
s = r + 2 b) Equation 1 is in slope-intercept form:
2 • \(\frac{1}{2}\)r = 2 • (\(\frac{1}{2}\)s – 1) We write Equation 2 in slope-intercept form:
r = s – 2
r + 2 = s – 2 + 2
s = r + 2
The two equations are equivalent therefore the system of equations is dependent and it has an infinity of solutions.

c) Comment on Ms. Cohen’s riddle.
Answer:
The answer ‘false’ to the question aims the situation in which the system is inconsistent while the answer ‘true’ reflects the cases in which the system either has a solution, or is dependent

a)
s = r + 2
\(\frac{1}{2}\)r = \(\frac{1}{2}\)s – 1
b) dependent system
c)The riddle is true as the system nas an infinity of solutions.

Question 15.
Math Journal Mr. Braunstein uses a cell phone plan that charges 2¢ per minute for local calls and 3.5¢ per minute for long distance calls. He made x minutes of local calls and y minutes of long distance calls in June. In July, he made 2x minutes of local calls and 2y minutes of long distance calls. His bill was $12 in June and $24 in July. Can he find how many minutes of each type of call he made each month? Explain.
Answer:
0.02x + 0.035y = 12 Equation 1
0.02(2x) + 0.035(2y) = 24 Equation 2
We write the system of equations:
All the coefficients of the two equations and the constants on the right are proportional, therefore the system is dependent which means the system has infinitely many solutions. This means he cannot find how many minutes of each type of call he made each month.
No (dependent system of equations)

Brain @ Work

Question 1.
Lorraine has $110 and Jane has $600 in their bank accounts. Lorraine’s account balance increases by $30 every year and her account balance will be C dollars in x years. Jane’s account balance reduces by $40 every year and her account balance will also be C dollars in x years.

a) Write two equations of C in terms of x.
Answer:
C = 110 + 30x
a) We write an equation for Lorraine account:
C = 600 — 40x We write an equation for Janes account:

b) Solve this system of linear equations to find the amount in the girls’ account balances when they are equal.
Answer:
C = 110 + 30x Equation 1 b) We write the system of equations:
C = 600 — 40x Equation 2
110 + 30x = 600 — 40x We substitute Equation 1 in Equation 2 and determine x:
110 + 30x + 40x = 600 — 40x + 40x
110 + 70x = 600
110 + 70x — 110 = 600— 110
70x = 490
x = \(\frac{490}{70}\)
x = 7
C = 110 + 30(7) Substitute 7 for c in Equation 1 to determine C:
= 110+ 210
= 320
a) C = 110 + 30x ; C = 110 + 30x
b) $320

Question 2.
The diagram below shows a 1-centimeter block of metals A and B.

Five blocks of A and two blocks of B have a total mass of 44 grams. Three blocks of A and five blocks of B have a total mass of 34 grams. An alloy is made by melting and mixing two blocks of metal A and one block of metal B.

Answer:
Lets note:
a = the mass of an A bIock
b = the mass of a B block
5a + 2b = 44 Equation 1
3a + 5b = 34 Equation 2
We write the system of equations:
5(5a + 2b) 5(44) Multiply Equation 1 by 5:
25a + 10b = 220 Equation 3
2(3a + 5b) = 2(34) Multiply Equation 2 by 2:
6a + 10b = 68 Equation 4 from Equation 3:
(25a + 10b) – (6a + 10b) = 220 — 68 we subtract Equation 4 from Equation 3:
25a + 10b — 6a — 1ob = 152 Simplify:
19a = 152
\(\frac{19 a}{19}\) = \(\frac{152}{19}\) We divide both sides by 19 and simplify:
a = 8
5(8) + 2b = 44 Substitute 8 for a in Equation 1 to determine b:
40 + 2b = 44
40 + 2b — 40 = 44 — 40
2b = 4
\(\frac{2 b}{2}\) = \(\frac{4}{2}\)
b = 2

we determine the density of the alloy
= \(\frac{2(8)+2}{3}\) = \(\frac{18}{3}\)
= 6 grams/cm³

Question 3.
The table shows Joseph’s phone usage and the total charges over three months.

Joseph suspects that there are errors in the charges. Use systems of linear equations to check whether the charges are correct.
Answer:
Let’s note:
x = the cost of a minute of a local voice call
y = the cost of one minute of an cut-of-state voice call
\(\frac{60}{40}\) = \(\frac{30}{20}\) ≠ \(\frac{45}{34}\) We notice that then coefficients of the variables for January and March are proportional, but the constants are not:
60x + 30y = 45 Equation 1
40x + 20y = 34 Equation 2
This means that the system of equations corresponding to these months is inconsistent
Therefore there is no solution, which is not correct: the charges for at least one of the months January and March is wrong
Incorrect

Practice the problems of Math in Focus Grade 8 Workbook Answer Key Chapter 5 Review Test to score better marks in the exam.

Math in Focus Grade 8 Course 3 A Chapter 5 Review Test Answer Key

Concepts and Skills

Solve each system of linear equations using the elimination method.

Question 1.
3x + 2y = 18
2x + 3y = 22
Answer:
3x + 2y = 18 Equation 1
2x + 3y = 22 Equation 2
We are given the system of equations:
2 ∙ (3x + 2y) = 2 ∙ 18 We can eliminate first either x, or y.
6x + 4y = 36 Equation 3 We usually choose to eliminate the variable which has smaller coefficients, in this case both have the same coefficients. We choose to eliminate x.
Multiply Equation 1 by 2:
3 ∙ (2x + 3y) = 3 ∙ 22 Multiply Equation 2 by 3:
6x + 9y = 66 Equation 4
(6a + 9y — (6x + 4y) = 66 — 36 We subtract Equation 3 from Equation 4:
6x + 9y — 6x — 4y = 30 Use the distributive property:
5y = 30 Simplify, 6x is eliminated:
\(\frac{5 y}{5}\) = \(\frac{30}{5}\) Divide by 5 and simplify:
y = 6
3x + 2(6) = 18 Substitute 6 for y in Equation 1:
3x + 12 = 18 Simplify:
Subtract 12 from both sides:
3x + 12 — 12 = 18 — 12
3x = 6 Simplify:
\(\frac{3 x}{3}\) = \(\frac{6}{3}\) Divide by 3:
x = 2
y = 6 The solution to the system of linear equations is
x = 2
y = 6

Question 2.
5x + y = 8
x + 3y = 10
Answer:
5x + y = 8 Equation 1 We are given the system of equations:
x + 3y = 10 Equation 2
3 ∙ (5x + y) = 3 ∙ 8 We can eliminate first either x, or y.
15x + 3y = 24 Equation 3 We usually choose to eliminate the variable which has smaller coefficients, in this case y. Multiply Equation 1 by 3:
(15x + 3y) – (x + 3y) = 24 – 10 We subtract Equation 2 from Equation 3:
15x + 3y – x – 3y = 14 Use the distributive property:
14x = 14 Simplify, 3y is eliminated:
\(\frac{14 x}{14}\) = \(\frac{14}{14}\)
Divide by 14 and simplify:
x = 1
5(1) + y = 8 Substitute 1 for x in Equation 1:
5 + y = 8 Simplify:
5 + y – 5 = 8 – 5 Subtract 5 from both sides:
y = 3 Simplify:
x = 1
The solution to the system of linear equations is
y = 3
x = 1; y = 3

Question 3.
\(\frac{1}{2}\)a + b = 7
a + 3b = 19
Answer:
\(\frac{1}{2}\)a + b = 7 Equation 1
a + 3b = 19 Equation 2
We are given the system of equations:
2 ∙ (\(\frac{1}{2}\)a + b) = 7 We can eliminate rirst either a, or b
a + 2b = 14 Equation 3 We usually choose to etiminate the variable which has smaller coefficients, in this case a.
Multiply Equation 1 by 2:
(a + 3b) — (a + 2b) = 19 — 14 We subtract Equation 3 from Equation 2:
a + 3b — a — 2b = 5 Use the distributive property
b = 5 Simplify, a is eliminated:
a + 3(5) = 19 Substitute 5 for b in Equation 2:
a + 15 = 19 Simplify:
a + 15 — 15 = 19 — 15 Subtract 15 from both sides:
a = 4 Simplify:
a = 4 The solution to the system of linear equations is
b = 5
a = 4 ; b = 5

Solve each system of linear equations using the substitution method.

Question 4.
a + 2b = 1
2a + b = 8
Answer:
a + 2b = 1 Equation 1 We are given the system of equations:
2a + b = 8 Equation 2
2a + b = 8 Use Equation 2 to express b in terms of a:
2a + b — 2a = 8 — 2a Subtract 2a from both sides:
b = 8 — 2a Equation 3 Simplify:
a + 2(8 — 2a) = 1 Substitute Equation 3 into Equation 1:
a + 16 — 4a = 1 Use the distributive property:
—3a + 16 = 1 Simplify:
—3a + 16 — 16 = 1 — 16 Subtract 16 from both sides:
—3a = —15 Simplify:
\(\frac{-3 a}{-3}\) = \(\frac{-15}{-3}\) Divide by -3:
a = 5 Simplify:
b = 8 — 2(5) Substitute 5 for a in Equation 3:
= 8 — 10
= -2
a = 5 The solution of the system ¡S:
b = -2
a = 5; b = -2

Question 5.
2x + 11y = 15
x – y = 1
Answer:
2x + 11y = 15 Equation 1 We are given the system of equations:
x — y = 1 Equation 2
x — y = 1 Use Equation 2 to express x in terms of y:
x – y + y = 1 + y Add y to both sides:
x = y + 1 Equation 3 Simplify:
2(y + 1) + 11y = 15 Substitute Equation 3 into Equation 1:
2y + 2 + 11y = 15 Use the distributive property:
13y + 2 = 15 Simplify:
13y + 2 — 2 = 15 — 2 Subtract 2 from both sides:
13y = 13 Simplify:
\(\frac{13 y}{13}\) = \(\frac{13}{13}\) Divide by 13:
y = 1 Simplify:
x = 1 + 1
= 2 Substitute 1 for y in Equation 3:
x = 2 The solution of the system is:
y = 1
x = 2
y = 1

Question 6.
\(\frac{1}{2}\)x + \(\frac{1}{2}\)y = 7
3x – y = 22
Answer:
We are given the system of equations:
\(\frac{1}{2}\)x + \(\frac{1}{2}\)y = 7 Equation 1
3x — y = 22 Equation 2
3x — y = 22 use Equation 2 to express y is terms of x:
3x — y — 3x = 22 — 3x Subtract 3x from both sides:
—y = 22 — 3x Simplify:
y = 3x — 22 Equation 3
\(\frac{1}{2}\)x + \(\frac{1}{2}\)(3x – 22) Substitute Equation 3 into Equation 1:
2 • \(\frac{1}{2}\)x + 2 • \(\frac{1}{2}\)(3x —22) = 2 • 7 Multiply both sides by 2:
x + 3x — 22 = 14 Simplify:
4x — 22 = 14
4x — 22 + 22 = 14 + 22 Add 22 to both sides:
4x = 36 Simplify:
\(\frac{4 x}{4}\) = \(\frac{36}{4}\) Divide by 4
x = 9 Simplify:
y = 3(9) — 22 Substitute 9 for x in Equation 3:
= 27 — 22
= 5
x = 9 The solution of the system is:
y = 5
x = 9
y = 5

Solve each system of linear equations using the graphical method. Use 1 grid square to represent 1 unit on both axes for the interval -3 to 8.

Question 7.
2x + 3y = 24
2x + y = 12
Answer:
2x + 3y = 24 Equation 1
2x + y = 12 Equation 2
We are given the system of equations:
2x + 3y = 24 We rewrite Equation 1 in slope-intercept form:
2x + 3y – 2x = 24 — 2x
3y = —2x + 24
\(\frac{3 y}{3}\) = \(\frac{-2 x+24}{3}\)
y = –\(\frac{2}{3}\)x + 8
2x + y = 12 we rewrite Equation 2 in slope-intercept rorm:
2x + y – 2x = 12 — 2x
y = —2x + 12

We graph the two equations: the first has slope —\(\frac{2}{3}\) and y-intercept 8, while the second has slope -2 and y-intercept 12:
x = 3 The point of intersection is (3, 6). Therefore the solution of the system of equations is
y = 6
x = 3; y = 6

Question 8.
2x + 5y = 1
3x – y = – 7
Answer:
2x + 5y = 1 Equation 1
3x – y = —7 Equatiou 2
We are given the system of equations:
2x + 5y = 1 We rewrite Equation 1 in slope-intercept form:
2x + 5y — 2x = 1 — 2x
5y= —2x + 1
\(\frac{5 y}{5}\) = \(\frac{-2 x+1}{5}\)
y = –\(\frac{2}{5}\)x + \(\frac{1}{5}\)
3x — y = —7 We rewrite Equation 2 in slope-intercept form:
3x — y — 3x = —7 — 3x
—y = —3x — 7
y = 3x + 7

We graph the two equations: the first has slope —\(\frac{2}{5}\) and y-intercept \(\frac{1}{5}\), while the second has slope 3 and y-intercept 7:
x = -2 The point of intersection is (—2, 1). Therefore the solution of the system of equations is:
y = 1
x = —2; y = 1

Question 9.
x + 0.5y = 6
3x – y = 13
Answer:
x + 0.5y = 6 Equation 1
3x — y = 13 Equation 2
We are given the system of equations:
x + 0.5y = 6 We rewrite Equation 1 in stope-intercept form:
x + 0.5y — x = 6 — x
0.5y = —x + 6
\(\frac{0.5 y}{0.5}\) = \(\frac{-x+6}{0.5}\)
y = -2x + 12
3x — y = 13 We rewrite Equation 2 in slope-intercept form:
3x — y — 3x = 13 — 3x
—y = —3x + 13
y = 3x — 13

We graph the two equations: the first has slope -2 and y-intercept 12. while the second has slope 3 and y-intercept -13:
x = 5 The point of intersection is (5, 2). Therefore the solution of the system of equations is
y = 2
x = 5; y = 2

Solve each system of linear equations. Explain your choice of method.

Question 10.
x = 4y – 1
2x – 6y = – 1
Answer:
x = 4y — 1 Equation 1 We are given the system of equations:
2x — 6y = —1 Equation 2
5x + y = 15 As there is a variable with coefficient 1 (which is x in Equation 1), we can use the substitution method.
x = 4y — 1 Equation 1 expresses x is terms of y:
2(4y – 1) – 6y = —1 Substitute Equation 1 into Equation 2:
8y — 2 – 6y = —1 Use the distributive property
2y — 2 = —1 Simplify:
2y — 2 + 2 = —1 + 2 Add 2 to both sides:
2y = 1 Simplify:
\(\frac{2 y}{2}\) = \(\frac{1}{2}\) Divide by 2:
y = \(\frac{1}{2}\) simplify
x = 4(\(\frac{1}{2}\)) – 1 Substitute \(\frac{1}{2}\) for y in Equation 1:
= 2 – 1
= 1
x = 1 The solution of the system is
y = \(\frac{1}{2}\)
x = 1
y = \(\frac{1}{2}\)

Question 11.
3x – 14y = -49
5x + 2y = 45
Answer:
3x — 14y = -49 Equation 1
5x + 2y = 45 Equation 2
We are given the system of equations:
7 • (5x + 2y) = 7 . 45 The substitution method would lead to fractions as none of the variables has coefficient 1.
35x + 14y = 315 Equation 3
We use the elimination method.
We can eliminate first either z, or g.
We choose to eliminate the variable y because -14 is multiple of 2.
Multiply Equation 2 by 7:
(3x — 14y) + (35x + 14y) = —49 + 315 We add Equation 3 to Equation 1:
3x — 14y + 35x + 14y = 266 Simplify, 14y is eliminated:
38x = 266
\(\frac{38 x}{38}\) = \(\frac{266}{38}\) Divide by 38 and simplify
x = 7
5(7) + 2y = 45 Substitute 7 for x in Equation 2:
35 + 2y = 45 Simplify:
35 + 2y — 35 = 45 — 35 Subtract 35 from both sides:
2y = 10 Simplify:
\(\frac{2 y}{2}\) = \(\frac{10}{2}\) Divide by 2
y = 5 simplify
x = 7 The solution to the system of linear equations is
y = 5
x = 7
y = 5

Question 12.
x – \(\frac{1}{3}\)y = 6
2x + y = 2
Answer:
x – \(\frac{1}{3}\)y = 6 Equation 1
2x + y = 2 Equation 2
We are given the system of equations:
x – \(\frac{1}{3}\)y = 6 We can use substitution or graphical method as we have at least one variable with coefficient 1. We choose the graphical method
x – \(\frac{1}{3}\)y – x = 6 – x
–\(\frac{1}{3}\)y = -x + 6 We rewrite Equation 1 in slope-intercept form:
—3(\(\frac{1}{3}\)y) = —3(—x + 6)
y = 3x— 18
2x + y = 2 We rewrite Equation 2 in slope-intercept form:
2x + y — 2x = 2— 2x
y = —2x + 2

We graph the two equations: the first has slope 3 and y-intercept -18, while the second has slope -2 and y-intercept 2:
x = 5 The point of intersection is (5, 2). Therefore the solution of the system of equations is
y = 2
x = 5
y = 2

Identify whether each system of equations is inconsistent, dependent, or has a unique solution. Justify your answer. Solve the system of linear equations if it has a unique solution.

Question 13.
3x + 2y = 8
6x + 4y = 16
Answer:
3x + 2y = 8 Equation 1
6x + 4y = 16 Equation 2
we are given the system of equations
3x + 2y = 8 Rewrite Equation 1 in slope-intercept form y = mx + b
3x + 2y – 3x = 8 – 3x Subtract 3x from both sides and simplify
2y = -3x + 8
\(\frac{2 y}{2}\) = \(\frac{-3 x+8}{2}\) Divide both sides by 2 and simplify
y = \(-\frac{3}{2} x\) + 4
6x + 4y = 16 Rewrite Equation 2 in slope-intercept form y = mx + b:
6x + 4y — 6x = 16 — 6x Subtract 6x from both sides and simplify:
4y = —6x + 16
\(\frac{4 y}{4}\) = \(\frac{-6 x+16}{4}\) Divide both sides by 4 and simplify:
y = \(-\frac{3}{2} x\) + 4
Since Equation 1 and Equation 2 are equivalent, they have an infinite number of solutions.
Therefore system of linear equations is dependent
Dependent

Question 14.
\(\frac{x}{3}\) + y = 8
x + 3y = 10
Answer:
\(\frac{1}{3} x\) + y = 8 Equation 1
x + 3y = 10 Equation 2
We are given the system of equations:
\(\frac{1}{3} x\) + y = 8
Rewrite Equation 1 in slope-intercept form y = mx + b:
\(\frac{1}{3} x\) + y – \(\frac{1}{3} x\) = 8 – \(\frac{1}{3} x\) Subtract \(\frac{1}{3} x\) from both sides and simplify:
y = –\(\frac{1}{3} x\) + 8
x + 3y = 10 Rewrite Equation 2 in slope-intercept form y = mx + b
x + 3y — x = 10 — x Subtract x from both sides and simplify:
\(\frac{3 y}{3}\) = \(\frac{-x+10}{3}\) Divide both sides by 3 and simplify:
y = \(-\frac{1}{3} x\) + \(\frac{10}{3}\)
The linear equations have the same slope and different y-intercept. The system of linear equations is inconsistent because the system of Linear equations has no solution.
Inconsistent

Question 15.
\(\frac{1}{2}\)x + y = 7
x + 2y = 14
Answer:
\(\frac{1}{2}\)x + y = 7 Equation 1
x + 2y = 14 Equation 2
We are given the system of equations:
\(\frac{1}{2}\)x + y = 7 Rewrite Equation 1 in slope-intercept form y = mx + b:
\(\frac{1}{2}\)x + y — \(\frac{1}{2}\)x = 7 — \(\frac{1}{2}\)x Subtract \(\frac{1}{2}\)x from both sides and simplify:
y = –\(\frac{1}{2}\)x + 7
x + 2y = 14 Rewrite Equation 2 in slope-intercept form y = mx + b:
x + 2y — x = 14 — x Subtract x from both sides and simplify:
\(\frac{2 y}{2}\) = \(\frac{-x+14}{2}\) Divide both sides by 2 and simplify:
Since Equation 1 and Equation 2 are equivalent they have an infinite number of solutions.
Therefore system of linear equations is dependent.
Dependent

Question 16.
2x – \(\frac{1}{2}\)y = 10
y = 4x + 11
Answer:
2x — \(\frac{1}{2}\)y = 1o Equation 1
y = 4x + 11 Equation 2
We are given the system of equations:
2x – \(\frac{1}{2}\)y = 10 Rewrite Equation 1 in slope-intercept form y = mx + b:
2x – \(\frac{1}{2}\)y – 2x = 10 – 2x Subtract 2x from both sides and simplify:
–\(\frac{1}{2}\)y = -2x + 10
-2 ∙ (-\(\frac{1}{2}\)y) = -2(-2x + 10) Multiply both sides by -2 and simplify
y = 4x — 20
y = 4x + 11 Equation 2 is written in slope-intercept form y = mx + b
The linear equations have the same slope and different y-intercept The system of linear equations is inconsistent because the system of linear equations has no solution.
Inconsistent

Question 17.
2x – 5y = -21
4x + 3y = 23
Answer:
2x — 5y = —21 Equation 1
4x + 3y = 23 Equation 2
We are given the system of equations:
2(2x — 5y) = 2(—21) We use the elimination method.
4x — 10y = —42 Equation 3 Multiply Equation 1 by 2:
(4x + 3y) — (4x — 10y) = 23 — (—42) We subtract Equation 3 from Equation 2 and simplify:
4x + 3y — 4x + 10y = 65
13y = 65
\(\frac{13 y}{13}\) = \(\frac{65}{13}\) Divide by 13 and simplify:
y = 5
4x + 3(5) = 23 Substitute 5 for y in Equation 2:
4x + 15 = 23
4x + 15 — 15 = 23 — 15 Subtract 15 from both sides and simplify:
4x = 8
The system of linear equations has an unique solution:
\(\frac{4 x}{4}\) = \(\frac{8}{4}\) Divide by 4 and simplify:
x = 2
x = 2 The system of Linear equations has an unique solution:
y = 5
Unique solution (x = 2, y = 5)

Question 18.
6y – 12x = 60
4y – 40 = 8x
Answer:
6y — 12x = 60 Equation 1
4y — 40 = 8x Equation 2
We are given the system of equations:
6y — 12x = 60 Rewrite Equation 1 in slope-intercept form y = mx + b:
6y — 12x + 12x = 60 + 12x Add 12x to both sides and simplify
6y = 12x + 60
\(\frac{6 y}{6}\) = \(\frac{12 x+60}{6}\) Divide both sides by 6 and simplify:
y = 2x + 10
4y – 40 = 8x Rewrite Equation 2 in slope-intercept form y = mx + b
4y — 40 + 40 = 8x + 40 Add 40 to both sides and simplify:
4y = 8x + 40
\(\frac{4 y}{4}\) = \(\frac{8 x+40}{4}\) Divide both sides by 4 and simplify:
y = 2x + 10
Since Equation 1 and Equation 2 are equivalent, they have an infinite number of solutions. Therefore system of linear equations is dependent
Dependent

Problem Solving

Solve. Show your work.

Question 19.
Andy and Ben both worked a total of 88 hours one week. Ben worked 8 hours more than Andy. Find the number of hours each man had worked.
Answer:
Lets note:
a = the number of hours worked by Andy
b = the number of hours worked by Ben
a + b = 88 Equation 1 We write the system of equations:
b = a + 8 Equation 2
We use the substitution method.
b is expressed in terms of a in Equation 2:
a + (a + 8) = 88 Substitute Equation 2 into Equation 1:
a + a + 8 = 88 Simplify:
2a +8 = 88
2a + 8 – 8 = 88 — 8 Subtract 8 from both sides:
2a = 80 Simplify:
\(\frac{2 a}{2}\) = \(\frac{80}{2}\) Divide both sides by 2:
a = 40 simplify
b = 40 + 8 Substitute 40 for a in Equation 2:
= 48
So Andy worked 40 hours and Ben worked 48 hours
Andy: 40 hours
Ben: 48 hours

Question 20.
In three years’ time, Mr. Sullivan will be 3 times as old as his daughter. Six years ago, he was 6 times as old as she was. How old are they now?
Answer:
s + 3 = 3(d + 3) Equation 1
s — 6 = 6(d — 6) Equation 2
Lets note:
s = Mr Sullivans age now
d = the daughters age now
We can write the system of equations:
s — 6 = 6d — 36 We use the substitution method.
s — 6 + 6 = 6d — 36 + 6 Rewrite Equation 2 to express s in terms of d:
s = 6d — 30 Equation 3
(6d — 30) + 3 = 3(d + 3) Substitute Equation 3 in Equation 1 and simplify:
6d — 30 + 3 = 3d + 9
6d — 27 = 3d + 9
6d — 27 – 3d = 3d + 9 – 3d We subtract 3d from both sides:
3d — 27 = 9
3d — 27 + 27 = 9 + 27 Add 27 to both sides:
3d = 36
\(\frac{3 d}{3}\) = \(\frac{36}{3}\) Divide by 3 and simplify:
d = 12
s = 6(12) — 30 Substitute 12 for d in Equation 3:
= 72 – 30
= 42
d = 12 The solution to the system of linear equations is:
s = 42
Mr Sullivan’s age is 42 years, while his daughter’s is 12 years now
Mr Sullivan: 42 years
The daughter: 12 years

Question 21.
Samantha has a riddle for her sister: Find a pair of integers x and y that satisfy 7x + 3y = 64 and one of the integers is 3 times the other.
Answer:
7x + 3y = 64 Equation 1
x = 3y Equation 2
We write the systems of equations:
7x + 3y = 64 Equation 1
y = 3x Equation 2
x = 3y Equation 1 We solve the first system using substitution.
7(3y) + 3y = 64 Substitute Equation 2 into Equation 1:
21g + 3y = 64
24y = 64
\(\frac{24 y}{24}\) = \(\frac{64}{24}\) We determine y:
y = \(\frac{8}{3}\)
As y is not integer, this is not a solution.
y = 3x Equation 2 We solve the second system using substitution.
7x + 3(3x) = 64 Substitute Equation 2 into Equation 1:
7x + 9x = 64
16x = 64
\(\frac{16 x}{16}\) = \(\frac{64}{24}\) We determine x:
x = 4
y = 3(4) Substitute 4 for x in Equation 2:
= 12
x = 4 The two integers are 4 and 12.
y = 12
x = 4
y = 12

Question 22.
The shape ABC in the quilt block below is an isosceles triangle. In triangle ABC, AB = AC. The perimeter of triangle ABC is 27.3 inches. Find the values of x and y. Then find the length of each side of the triangle in inches.

Answer:
x + 5 = 6x — 2y Equation 1
2(x + 5) + y + 6.3 = 27.3 Equation 2 We can write the system of equations:

x + 5 – x = 6x — 2y — x Equation 1
2x + 10 + y + 6.3 = 27.3 Equation 2
we bring me system to the standard form:

5x — 2y = 5 Equation 1
2x + y + 16.3 = 27.3 Equation 2

5x — 2y = 5 Equation 1
2x + y + 16.3 — 16.3 = 27.3 — 16.3 Equation 2

5x — 2y = 5 Equation 1
2x + y = 11 Equation 2

2x + y = 11 We use the substitution method.
2x + y — 2x = 11 — 2x We express y in terms of x from Equation 2:
y = —2x + 11 Equation 3
5x — 2(—2x + 11) = 5 Substitute Equation 3 into Equation 1:
5x + 4x — 22 = 5
9x — 22 = 5
9x — 22 + 22 = 5 + 22 We determines x:
9x = 27
\(\frac{9 x}{9}\) = \(\frac{27}{9}\)
x = 3
y = —2(3) + 11 Substitute 3 for x in Equation 3:
= -6 + 11
= 5
x = 3 The system’s solution is:
y = 4
AB = AC = x + 5 = 3 + 5 = 8 We determine AB and AC:
BC = y + 6.3 = 5 + 6.3 = 11.3 We determine BC:
x = 3; y = 5
AB = 8 inches; AC = 8 inches BC = 11.3 inches

Question 23.
A bus company requires 4 buses and 8 vans to take 240 school children to the library. It requires 2 buses and 9 vans to take 170 children to the museum. Calculate the number of children a bus can carry and the number of children a van can carry.
Answer:
Let’s note:
b = the number of children in a bus
v = the number of children in a van
4b + 8v = 240 Equation 1
26 + 9v = 170 Equation 2
We write the system of equations:
2(2b + 9v) = 2(170) We use the elimination method.
4b + 18v = 340 Equation 3 Multiply Equation 2 by 2:
(4b + 18v) — (4b + 8v) = 340 — 240 Subtract Equation 1 from Equation 3:
4b + 18v — 4b — 8v = 100 Simplify:
10v = 100
\(\frac{10 v}{10}\) = \(\frac{100}{10}\) Divide by 10 and simplify:
v = 10
2b + 9(10) = 170 Substitute 10 for v in Equation 2:
2b + 90 = 170
2b + 90 — 90 = 170 — 90 Subtract 90 from both sides:
2b = 80
\(\frac{2 b}{2}\) = \(\frac{80}{2}\) Divide both sides by 2:
b = 40 Simplify:
A bus can carry 40 children, while a van can carry 10 children.
Bus: 40 children
Van : 10 children

Question 24.
At noon, Balloon M is 60 meters above ground, and Balloon N is 50 meters above ground. Balloon M is rising at the rate of 10 meters per second, while Balloon N is rising at the rate of 15 meters per second.

a) Write a system of two linear equations in which each equation gives the height, h meters, of a balloon t seconds after noon. Then solve the system using a graphing calculator.
Answer:
h = 60 + 10t Equation 1 a) We write a system of equations giving the height of each balloon:
h = 50 + 15t Equation 2

We use a graphing calculator to solve the system:
t = 2 The solution is:
h = 80

b) How many seconds after noon will the two balloons be at the same height?
How do you know?
Answer:
t = 2 seconds b) The two balloons will be at the same height when h = 80. We determined the time when this happens, which is the t-coordinate of the solution above:
a) t = 2, h = 80
b) 2 seconds

Question 25.
The water levels in two identical tanks rise at a rate of 4 inches per second. The water level, h inches, in Tank A is 3 inches at t = 0. The water level in Tank B is 16 inches at t = 3 seconds.

a) Write a system of equations for the water level h in the two tanks in terms of t.
Answer:
h = 3 + 4t Equation 1 a) we write system of equations for the water level in the two tanks
h = 16 + 4(t – 3) Equation 2

b) Graph the two equations on the same coordinate plane. Use 1 grid square on both axes to represent 1 unit for the interval 0 to 10.
Answer:
h = 4t + 3 Equation 1 b) We write the equations in slope-Intercept form
h = 16 + 4t – 12 Equation 2
h = 4t + 3 Equation 1
h = 4t + 4 Equation 2

We graph the two equations the first has slope 4 and y-Intercept 3, while the second has slope 4 and y-intercept 4.

c) When will the two tanks have the same water level? How do you know?
Answer:
The two tanks will never have the same level because the system of equations in part (a) is inconsistent because the slope is the same, but the y-intercept is not

a)
h = 3 + 4t
h = 16 + 4(t — 3)
b) See graph
c) Never

Question 26.
Natalie has x bags of onions and y bags of potatoes. There are a total of 8 bags. Each bag weighs 2 pounds. The total weight of the bags is 16 pounds. She wants to find the value of x and y.

a) Write a system of two linear equations.
Answer:
x + y = 8 Equation 1
2x + 2y = 16 Equation 2
a) We write a system of equations:

b) State with reasons whether the system of equations has a unique solution, is inconsistent, or is dependent.
Answer:
\(\frac{1}{2}\) = \(\frac{1}{2}\) = \(\frac{8}{16}\) b) We notice that the coefficients of the variables and the constants are proportional:
This means that the system is dependent

c) Can Natalie find the value of x and y? Why?
Answer:
She cannot find the value of x and y because there are an infinity of solutions.

a)
x + y = 8
2x + 2y = 16
b) Dependent system
c) No

This handy Math in Focus Grade 8 Workbook Answer Key Chapter 9 Congruence and Similarity detailed solutions for the textbook questions.

Math in Focus Grade 8 Course 3 B Chapter 9 Answer Key Congruence and Similarity

Math in Focus Grade 8 Chapter 9 Quick Check Answer Key

Complete.

Question 1.

Answer:

Explanation:
1. Scale factor: 2
2 × 8ft = 16 ft
2. Original length = 5
Scaled length = 5
Scale factor = 5/5 = 1
3. Original length = 8 m
Scaled length = 4 m
Scale factor = scale/actual = 4/8 = 1/2
4. Scale factor = 1/4
Original length = 4 inches
We know that
scale factor = scale/actual
1/4 = scale length/4
scaled length = 4/4
scaled length = 1 in.

Solve.

Question 2.
Shawn built a model of a ship whose length was 4,500 inches. His model was 15 inches long. Find the scale factor.
Answer:
Let the scale factor be x
Shawn built a model of a ship whose length was 4,500 inches.
His model was 15 inches long.
x × 15 = 4500
x = 4500/15
x = 300 inches

Question 3.
A line segment is 4 centimeters long. When projected on a screen, the line segment is 20 centimeters long. Find the scale factor.
Answer:
Let the scale factor be x
scale factor = scale/actual
Actual length = 4 cm
scaled length = 20 cm
scale factor = 20 cm/4 cm
scale factor = 5 cm
Thus the scale factor is 5 cm.

Question 4.
A model plane is built with a scale factor of \(\frac{1}{180}\). The actual length of the plane is 210 feet (2,520 inches). Find the length of the model.
Answer:
Given,
A model plane is built with a scale factor of \(\frac{1}{180}\).
The actual length of the plane is 210 feet (2,520 inches).
Scale/Actual = 1/180
1/180 = l/210 feet
210/180 = length
Length = 210/180
Length = 21/18
Length = 7/6
Length = 1.16 feet

Question 5.
The scale of a map is 2 inches : 3 kilometers. The length of a road on the map is 3 inches. Find the actual length of the road.
Answer:
Given,
The scale of a map is 2 inches : 3 kilometers.
The length of a road on the map is 3 inches
Let the actual length of the road be x.
2 in./ 3 km = 3 in / x km
x = (3 × 3)/2
x = 9/2 = 4.5 km
Thus the actual length of the road is 4.5 kilometers.

Question 6.
The diagram shows a plot of land ABCD drawn on a map.

a) On the map, \(\overline{C D}\) is 1\(\frac{3}{4}\) inches. Find the scale of the map.
Answer:

b) Find the area of ABCD on the map.
Answer:

The measures of two interior angles are given for each triangle. Find the measure of the third interior angle.

Question 7.
△ABC: 20°, 80°
Answer:
Given
△ABC: 20°, 80°
The sum of interior angles of the triangle = 180°
∠a + ∠b + ∠c = 180°
20°+ 80° + ∠c = 180°
100° + ∠c = 180°
∠c = 180° – 100°
∠c = 80°
Thus the measure of the third interior angle is 80°

Question 8.
△KLM: 37°, 76°
Answer:
Given,
△KLM: 37°, 76°
The sum of interior angles of the triangle = 180°
∠k + ∠l + ∠m = 180°
37°+ 76° + ∠m = 180°
113° + ∠m = 180°
∠m = 180° – 113°
∠m = 67°
Thus the measure of the third interior angle is 67°

Question 9.
△PQR: 15°, 103°
Answer:
Given,
△PQR: 15°, 103°
The sum of interior angles of the triangle = 180°
∠p + ∠q + ∠r = 180°
15°+ 103° + ∠r = 180°
∠r = 180° – 118°
∠r = 62°

Find the unknown angle measure.

Question 10.

Answer:

Let the unknown angle be y
The sum of three angles = 180°
20° + 27° + y = 180°
47° + y = 180°
y = 180° – 47°
y = 133°
x° + y° + 27° = 180°
x° + 133°  + 27° = 180°
x° = 180° – 133° – 27°
x° = 20°
Thus the unknown angle is 20°

Question 11.

Answer:
y° + y° = 120°
2y° = 120°
y = 120°/2
y =60°

Solve for each variable.

Question 12.

Answer:
a° + 2b° = 180
b° = 40° (alternate angles)
2b° = 80°
a° + 2b° = 180
a° + 80° = 180°
a° = 180° – 80°
a° = 100°

Question 13.

Answer:
c° + 28° = 180°
c° = 180° – 28°
c° = 152°

Practice the problems of Math in Focus Grade 8 Workbook Answer Key Chapter 8 Review Test to score better marks in the exam.

Math in Focus Grade 8 Course 3 B Chapter 8 Review Test Answer Key

Concepts and Skills

State whether a rotation, translation, or a combination of both is involved in each activity.

Question 1.
A turning blade of a windmill
Answer:
A turning blade of a windmill is rotation. Because the blade of a windmill rotates in a horizontal axis.

Question 2.
Pressing the keys on a computer keyboard
Answer:
Pressing the keys on a computer keyboard is a translation. Because the keyboard translates the number into binary data.

Question 3.
A printer head moving left and right
Answer:
A printer head moving left and right is a translation. Because it is moving from left to right and right to left it means translating from left to right and right to left.

Question 4.
Wheels on a moving bicycle
Answer:
Wheels on a moving bicycle is a rotation. Because the bicycle wheel is rotated in clockwise and anticlockwise.

Describe the translations.

Question 5.
Climbing up 8 steps of a staircase (assume horizontal and vertical distances of each step are the same)
Answer:
The translation is a type of transformation that moves each point in a figure with the same distance in the same direction. Climbing up 8 steps of a staircase is a translation.

Question 6.
Taking an elevator from level 2 to level 5 of a building
Answer: Taking an elevator from level 2 to level 5 of a building is translation. Because the elevator is a device that moves up and down.

Write an equation of the line(s) of reflection.

Question 7.

Answer:
Reflection in x axis is (x,y) = (x,-y)
Reflection in y axis is (x,y) = (-x,y)
In the above figure consider the left image as ABCD and right image as PQRS
The reflection in y axis is
P(1,1) reflection is A(-1,1)
Q(3,1) reflection is B(-3,1)
R(4,-1) reflection is C(-4,-1)
S(2,-1) reflection is D(-2,-1)

Question 8.

Answer:
Reflection in x axis is (x,y) = (x,-y)
Reflection in y axis is (x,y) = (-x,y)
Consider the image below the x axis is OAB and above the x axis is OPQ
O(0,0) reflection is O(0,0)
A(0,-1) reflection in x axis is P(0,-(-1)) = P(0,1)
B(-1,0) reflection in y axis is Q(-(-1),0) = B(1,0)

Question 9.

Answer:
Reflection in x axis is (x,y) = (x,-y)
Reflection in y axis is (x,y) = (-x,y)
Consider the image below the x axis is ABC and above the x axis is PQR
A(1,1) reflection to x axis is P(1,-1)
B(0,3) reflection to x axis is Q(0,-3)
C(-2,1) reflection to x axis is R(-2,-1)

Question 10.

Answer:

Each diagram shows a figure and its line of reflection. On a copy of the graph, draw the image.

Question 11.

Answer:
Reflection in x axis is (x,y) = (x,-y)
Reflection in y axis is (x,y) = (-x,y)
Consider the image below the x axis is ABCD.
A(-3,-3) reflection to the x axis is (-3,3)
B(0,-3) reflection to the x axis is (0,3)
C(-1,-4) reflection to the x axis is (-1,4)
D(-2,-4) reflection to the x axis is (-2,4)

Question 12.

Answer:

Solve on graph paper. Show your work.

Question 13.
\(\overline{\mathrm{AB}}\) is dilated with center at the origin and scale factor 2. Draw \(\overline{\mathrm{AB}}\) and its image \(\overline{A^{\prime} B^{\prime}}\). Use 1 grid square on the horizontal axis to represent 1 unit for x interval from -4 to 10, and 1 grid square on the vertical axis to represent 1 unit for the y interval from -2 to 4.
a) A (2, 1) and B (5, 2)
Answer:

b) A (1, -1) and B (-2, 2)
Answer:

Question 14.
The table shows the coordinates for ∆XYZ and its images using two transformation. Use 1 grid square on both axes to represent 1 unit for the interval from -3 to 9.

a) ∆XYZ is mapped onto ∆X’Y’Z’ and ∆X”Y”Z” by a dilation. Draw each triangle and its image on the same coordinate plane. Then mark and label D as the center of dilation.
Answer:

In the drawn graph ABC = XYZ, DEF = X’Y’Z’, GHI = X”Y”Z”

b) ∆XYZ is mapped onto ∆ABC by a rotation 90° counterclockwise about the origin. Draw ∆ABC on the coordinate plane.
Answer:

In the drawn map XYZ = ABC and ABC = DEF
90⁰ counterclockwise rotation = (-y,x)
X(1,3) = X(-3,1)
Y(1,1) = Y(-1,1)
Z(2,1) = Z(-1,2)

c) ∆XYZ is mapped onto ∆PQR by a translation 3 units to the left and 4 units up. Draw ∆PQR on the coordinate plane.
Answer:

In the drawn graph XYZ = ABC and PQR = DEF.
In the graph ∆DEF is the translation.

d) Compare the transformations that mapped ∆XYZ onto ∆ABC and ∆PQR in terms of preservation of the shape and size of ∆XYZ.
Answer:
The transformations ∆XYZ onto ∆ABC is rotation it means the figure turns in the clockwise or counter clockwise but it doesn’t change the shape and size.∆PQR onto ∆XYZ is a translation it means moving the figure on the coordinate plane without changing the shape and size.

Problem Solving

Solve. Show your work.

Question 15.
Jane had lunch with her friends from 1 P.M. to 2 P.M. Describe the geometric transformation of the hour hand of the clock.
Answer:

Question 16.
A scientist used a sensor to track the movement of a mouse. It moved from the point (-2, 3) to the point (8, 6). State the new coordinates of any point (x, y) under this translation.
Answer:
Given that,
The mouse moved from the point (-2,3) to the point (8,6)
(x1,y1) = (-2,3)
(x2,y2) = (8,6)
New coordinates are (x,y)
x coordinate = x2 – x1 = 8 – (-2) = 8 + 2 = 10.
y coordinate = y2 – y1 = (6-3) = 3.
Therefore, New coordinates are (x+10, y+3).

Question 17.
Mrs. Morales outlined a clover with four identical leaflets, as shown, on a coordinate plane. The center of the clover is at (1, 0).

a) How many lines of symmetry does the clover have? Sketch them on a copy of the clover leaf.
Answer:

b) Find an equation of each line of reflection.
Answer:
The equation of each line of reflection is x = 0, y = 0, y = -x and y = x
Therefore there are four symmetry.

Question 18.
A circular mold in a Petri dish had a diameter of \(\frac{1}{2}\) inch. The diameter grew by 32% in a day.

a) What is the scale factor of dilation?
Answer:
Given that the circular in a Petri dish has a diameter of ½ inch.
The diameter grew in a day = 32%
The formula for the scale factor of dilation is = dimensions of a new shape/dimensions of a old shape
= 32%/1/2 = 0.32/0.5 = 0.64

b) Find the diameter of the mold after a day.
Answer:
The circular mold in a Petri dish has a diameter of 1/2
The diameter grew in a day = 32%
= 2 × 0.32 = 0.64
Therefore the diameter of the mold after a day = 0.64.

Question 19.
A figurine of the Statue of Liberty is 10 inches tall. The height of the Statue of Liberty is 150 feet. What is the scale factor of the dilation if the figurine is the image of the statue?
Answer:
Given that,
Figurine of the statue of Liberty is 10 inches tall
The height of the statue of Liberty is 150 feet.
The scalar factor of the dilation if the figurine is the image of the statue = figurine of the statue of Liberty/height of the statue of Liberty
Here 1 feet = 12 inches
150 inches = 150 × 12 = 1800
= 10/1800
= 1/180
= 0.005
Therefore, the scale factor of the dilation is 0.005 inches.

Question 20.
A spotlight is placed 2 feet from a 1-foot tall vase. A shadow 5 feet tall is cast on a wall as shown in the diagram. Find the distance of the vase from the wall.

Answer:
Given that,
The spotlight is placed 2 feets away from a 1 foot tall vase.
A shadow 5 feet tall is cast on a Wall.
Consider the distance from the vase to the wall is d.
The distance of a vase from the wall = distance of shadow from the spot light/distance of vase from the spotlight
= 2+d/2 = 5
= 2+d = 10
d = 10 – 2
d = 8
Therefore the distance from the vase to the wall is 8 feet.

Practice the problems of Math in Focus Grade 8 Workbook Answer Key Chapter 8 Lesson 8.5 Comparing Transformations to score better marks in the exam.

Math in Focus Grade 7 Course 3 B Chapter 8 Lesson 8.5 Answer Key Comparing Transformations

Math in Focus Grade 8 Chapter 8 Lesson 8.5 Guided Practice Answer Key

Technology Activity

Materials:

• geometry software

Compare Transformations With Geometry Software

Step 1.
Draw two line segments intersecting at a common endpoint using a geometry software program. Call the angle formed ∠AOB.

Step 2.
Measure their lengths and the angles they make with the x-axis. Record your results in the “before” cells of the second column in the following table.

Step 3.
Translate the angle in both the horizontal and vertical directions. Complete the second column of the table.

Step 4.
Reflect the angle in either the x- or y-axis. Complete the third column of the table.

Step 5.
Rotate the angle about the origin. Complete the fourth column of the table.

Step 6.
Dilate the angle by any nonzero scale factor and with any center. Complete the last column of the table.

Step 7.
Note which transformations preserve angle measure. Why does this guarantee that perpendicular lines will be preserved? Why does this guarantee that parallel lines will be preserved?

Math Journal Compare translations, reflections, rotations, and dilations. Discuss which ones preserve lengths, angle measures, perpendicular lines, and parallel lines.

Solve on graph paper. Show your work.

Question 1.
∆ABC is mapped onto ∆PQR, ∆LMN, and ∆XYZ as shown on the coordinate plane. State which transformation was used to obtain each image. Justify your answers.

a) What transformation maps ∆ABC onto ∆PQR?
Answer:
The transformation maps ∆ABC onto the ∆PQR is translation.
Translation is a type of transformation that moves the figure on the coordinate plane without changing its orientation.
Here ∆ABC moves 2 units right through the x axis and it moves up 1 unit through the y axis.

b) What transformation maps ∆ABC onto ∆LMN?
Answer:
The transformation maps ∆ABC onto the ∆LMN is reflection.
Reflection means it is a type of transformation in which the image cutting out a preimage, lifting it, and putting it back face down is called reflection.

c) What transformation maps ∆ABC onto ∆XYZ?
Answer:
The transformation maps ∆ABC onto the ∆LMN is rotation.
Rotation means it is a type of transformation. It turns the image clockwise or counterclockwise but doesn’t change the image.
Here the image is rotated counterclockwise.

d) Compare the three transformations in terms of preserving lengths and angle measures.
Answer:
The three transformations are the transformation maps ∆ABC onto the ∆PQR is translation. After Changing ∆ABC onto the ∆PQR the length and angle measures are not changed.
The transformation maps ∆ABC onto the ∆LMN is reflection. After changing ∆ABC onto the ∆LMN the length and angle measures are not changed.
The transformation maps ∆ABC onto the ∆LMN is rotation. After Changing ∆ABC onto the ∆LMN the length is not changed but the angle measures are changed

You can use two vertices and their images to identify a transformation of one triangle onto another. But you should also use the third vertex and its image to check that other points of the figure are related by this transformation.

Math in Focus Course 3B Practice 8.5 Answer Key

Copy and complete on graph paper.

Question 1.
A curtain is made from a printed fabric with many pentagonal figures on it. One of the pentagons has coordinates A (1, 2), B (3, 2), C (4, 4), D (2, 6), and E (0, 4). Four other pentagons are images of ABCDE after each of the transformations in a) to d).
a) ABCDE is mapped onto FGHIJ by a translation of 8 units down. Draw FGHIJ.
Answer:
Given coordinates of Pentagon is
A (1, 2), B (3, 2), C (4, 4), D (2, 6), E (0, 4).
After translating the 8 units down the coordinates are F(1,-6), G(3,-6), H(4,-4), I(2,-2), J(0,-4).

b) ABCDE is mapped onto KLMNP by a reflection about the y-axis. Draw KLMNP.
Answer:
Given coordinates of Pentagon is
A (1, 2), B (3, 2), C (4, 4), D (2, 6), E (0, 4).
After reflection to the y axis the coordinates are K(1,-2), L(3,-3), M(4,-4), N(2,-6) P(0,-4)

c) ABCDE is mapped onto QRSTU by a rotation 90° clockwise about the origin, O. Draw QRSTU.
Answer:
Given coordinates of Pentagon is
A (1, 2), B (3, 2), C (4, 4), D (2, 6), E (0, 4).
After rotation of 90⁰ clockwise the coordinates are Q(-3,2), R(-1,2), S(0,-4), T(-2,6), U(-4,4).
And point the origin as O.

d) ABCDE is mapped onto VWXYZ by dilation with center at the origin and scale factor 1.5. Draw VWXYZ.
Answer:
Given that scale factor = 1.5
ABCDE is mapped onto VWXYZ by using the scale factor is 1.5 which means multiplying the coordinates with 1.5. then we get
A (1, 2) = V(1 × 1.5, 2 × 1.5) = (1.5,3)
B (3, 2) = W(3 × 1.5, 2 × 1.5) = (4.5,3)
C (4, 4) = X(4 × 1.5, 4 × 1.5) = (6,6)
D (2, 6) = Y(2 × 1.5, 6 × 1.5) = (3,9)
E (0, 4) = Z(0 × 1.5, 4 × 1.5) = (0,6)

e) Compare the four transformations in terms of lengths and angle measures of the pentagons.
Answer:
ABCDE is mapped onto FGHIJ by a translation of 8 units down. The length and angle measures of the pentagons are equal.
ABCDE is mapped onto KLMNP by a reflection about the y-axis. The length is the same but the angle measured is different from a pentagon.
ABCDE is mapped onto QRSTU by a rotation 90° clockwise about the origin. The length is the same but the angle is different from a pentagon.
ABCDE is mapped onto VWXYZ by a dilation with center at the origin and scale factor 1.5. The length and angle measures of a pentagon is different.

Question 2.
Quadrilateral P with coordinates (3, 2), (4, 5), (5, 5), and (6, 3) is mapped onto quadrilateral Q and R shown on the coordinate plane.
a) Describe the transformation from P to Q.
Answer:
The transformation from P to Q is translation. It means the figure on the coordinate plane moves without changing its orientation.

b) Quadrilateral P is mapped onto R by a half turn about the origin, O. Draw quadrilateral R.
Answer:
Note that a 180° rotation changes ttie coordinates from its original form (x, y) into (-x, -y). So, Figure P with points (3, 2), (4, 5), (5, 5) and (6, 3) will be mapped onto (-3,-2), (-4, -5), (-5, -5) and (-6, -3). Plotting these points then connecting it all determines image R.

c) The transformation that maps quadrilateral P onto quadrilateral R can be described in another way. Describe the transformation.
Answer:
The transformation that maps quadrilateral P onto quadrilateral R can be described in another way is reflection.
P with coordinates (3, 2), (4, 5), (5, 5), and (6, 3) after reflection the R coordinates are (3,-3), (6,-3), (5,-5), (4,-5).

d) Compare the transformations in a) to c) in terms of the preservation of the shape and size of quadrilateral P.
Answer :

Question 3.
There are three arrow signs J, K, and L on a street. Jennifer is standing at the origin, O.

a) Arrow J tells Jennifer where the Kodak Theatre is. Arrow K is the reflection of arrow J in the line y = 3. Label arrow J.
Answer:

Here arrow K is on the X-axis and the K is the reflection of arrow J in the y axis.

b) Jennifer sees arrow L that is an image of arrow J. Describe the transformation that maps arrow J onto arrow L.
Answer:
The arrow L is the image of J. The transformation that maps arrow J onto L is dilation.

c) Compare the two transformations in terms of the shape and size of the arrows.
Answer:
Here we are comparing the two transformations the image K and image L is the reflection, the shape and size are the same. The image L and image J are dilation the shape and size are different.

Solve on graph paper. Show your work.

Question 4.
Triangle ABC is mapped onto triangle A’B’C’ and triangle A”B”C” as shown on the coordinate plane.

a) Triangle ABC is mapped onto triangle A’B’C’ by a reflection in the line l. On a copy of the graph, draw line l and label its equation.
Answer:
Here draw the line along the x axis with the equation of I = 1.

b) Triangle ABC is mapped onto triangle A”B”C” by a dilation with scale factor -1. Mark and label T as the center of dilation on the coordinate plane in a).
Answer:
Scale factor -1 is multiplied by the Triangle A”B”C”. Then it forms the transformation is dilation. Dilation means the image is enlarged or reduced. And mark the center of dilation is T.

c) Describe another transformation that maps triangle ABC onto triangle A”B”C”.
Answer:
Here another transformation is the reflection. The triangle ABC is reflected as A”B”C”.

d) Compare the transformations that mapped triangle ABC onto triangle A’B’C’ and A”B”C” in terms of preservation of the shape and size of triangle ABC.
Answer:
The transformation of triangle ABC onto triangle A’B’C’ is the reflection and ABC onto triangle A”B”C” is the reflection and the dilation.
The transformation maps ∆ABC onto the ∆LMN is rotation.

Question 5.
∆ABC is mapped onto ∆A’B’C’ and ∆A”B”C”. The table of coordinates for the triangles is shown.

a) Draw ∆ABC, ∆A’B’C’, and ∆A”B”C”. Use 1 grid square on both axes to represent 1 unit for the interval from -5 to 7.
Answer:

b) ∆ABC is mapped onto ∆A’B’C’ by a dilation with center at the origin, O, and scale factor-1. Describe another transformation that maps ∆ABC onto ∆A’B’C’.
Answer:
Observe that the points for triangle ABC are on (-4, -3), (-3, -1), and (-2, -3) whiLe the points for triangle A’B’C’ are on (4, 3), (3, 1), and (2, 3). Also, note that a 180: rotation changes the coordinates from its original form (x, y) into (-x, -y), which is the case for mapping triangle A’B’C’ from the original triangle. Therefore, the transformation is a 180° rotation.

c) What is the transformation that maps ∆ABC onto ∆A”B”C”?
Answer:
The transformation that maps triangle ABC onto Triangle A”B”C” is the dilation it means the figure is enlarged or reduced.

The transformation that maps ∆ABC onto ∆A”B” C” is the dilation which means the figure is enlarged or reduced.

d) Compare the transformations that mapped ∆ABC onto ∆A’B’C’ and ∆A”B”C” in terms of preservation of the shape and size of ∆ABC.
Answer:
Notice that the reflection and rotation only moves the original triangle ABC into different Location and also changes its angular position. The original triangle if compared to the image (∆A’B’C’) that undergo these transformation still has the same measure of sides and angles even after the transformations. However, triangle ABC and A”B”C” are different in sizes. Notice that the dilation for this case makes the image bigger.

Question 6.
Math Journal How do you describe a transformation that maps an equilateral triangle onto itself using each of the following: translation, reflection, rotation, and dilation?
Answer:
The target of this task is to give the description of the rotation, dilation, translation and reflection of an equilateral triangle that maps itself to itself.

For rotation, note that a full rotation is 360°. Of course, every figure rotated at this degree maps itself to itself. Another angle of rotation for an equilateral triangle is a 120° rotation. Since all the sides and angles of the equilateral triangle are congruent then mapping one vertex of the triangle to another will just produce the same triangle in terms of size, position, and shape. Everything will be just congruent Therefore, a rotation at 120° makes the equilateral triangle and its image the same.

For dilation, note that the only scale factor that does not change orientation and size of any figure is 1. Therefore, the dilation at a scale of 1 makes the equilateral triangle and its image the same.

For the reflection, observe the altitude of the equilateral triangle. Creating a line of symmetry in this part of the triangle, divides the triangle into two equal parts. Thus, all the points have the same distance with its symmetric half from this Line. Therefore, the reflection of that makes the equiLateral triangle and its image the same is at the altitude of the triangle.

For the translation, there is no translation for an equilateral triangle that makes the triangle and its image the same. Since, if one vertex of the equilateral triangle moves up or down, or left or right all of its vertex will move that way, thus the whole triangle will move and change its location, unless the translation is 0.

Brain @ Work

Question 1.
A two-blade table fan is coded with two digits for easy identification. Each blade in the horizontal position shown in the picture is painted with the two-digit number. What number should be used so that when the blades make a half turn, the number will be read correctly?

Answer:
The goal is to determine the two-digit numbers that should e printed in the two-blade fan so that when it has rotated at an angle of 180° the number will not be upside down The trick is to choose two two-digit numbers for each blade that wherever the position of the blade is, horizontally, the number will stilt look the same. Consider numbers 11 and 69, at the original horizontal state of the blade, blade 11 will be at the right while blade 69 will be at the left After the rotation, blade 69 will be at the right while blade 11 will be at the left both can still be read correctly.

Question 2.
Line P with equation y = x + 2 maps onto line Q with equation y = -x + 4.

a) Describe a reflection that maps line P onto line Q.
Answer:

b) Describe a rotation that maps line P onto line Q.
Answer: Rotation means changing the figure clockwise or counterclockwise in the given equation line P rotation in the clockwise direction.

Question 3.
Triangle ABC is mapped onto triangle XYZ by a rotation 270° clockwise about center P. Both triangles are shown on the coordinate plane below. On a copy of the graph, mark and label the center P. What are the coordinates of center P? Explain how you found your answer.

Answer:
Plot the point P on the rotation 270° clockwise figure. The coordinates of the P are x(-1,1), y(-1,3) and z(-4,3).

Question 4.
After a rain, a circular puddle shrunk steadily for two days in the sun. The puddle was 8 inches in diameter initially. It shrunk by a scale factor each day. After two days, the diameter of the puddle was 6.48 inches. What is the scale factor of dilation? Explain how you found your answer.
Answer:
Given that the circular puddle shrunk steadily for two days in the sun then the diameter of a puddle is 8 inches.
After two days the diameter of a puddle is 6.48 inches.
Scale factor = Diameter of a changed puddle/ Diameter of an original puddle
= 6.48 inches/8 inches
= 0.81 inches
Therefore the scalar factor of a dilation is 0.81 inches.

Go through the Math in Focus Grade 4 Workbook Answer Key Chapter 3 Whole Number Multiplication and Division to finish your assignments.

Math in Focus Grade 4 Chapter 3 Answer Key Whole Number Multiplication and Division

Math Journal

Look at each problem. Use estimation to explain why the answers are not reasonable.

Example
5,268 × 8 = 2,144
Explain.
5,268 is about 5,000
5,000 × 8 = 40,000
So the answer is too small.

Question 1.
725 × 6 = 700
Explain.
Answer:
725 is about 700
700 × 6 = 4,200
So the answer is too small.

Question 2.
497 × 21 = 1,291
Explain.
Answer:
497 is about 500
500×21 = 10,500.
So the answer is too small.

Use estimation to explain why the answer is not reasonable.

Question 3.
6,021 ÷ 3 = 207
Explain.
____
Answer:
6,021 is about 6,000
6,000 ÷ 3 = 2,000.
So the answer is too small.

Solve. Show your work.

Question 4.
Look at the number sentence.
72 ÷ 6 = 12
How would you use this to find the missing quotient?
7200 ÷ 6 =

Answer:
7200 ÷ 6 = 1200.

Explanation:
As 72 ÷ 6 = 12, so for 7200 ÷ 6 it will be 1200.

Put on Your Thinking cap!

Challenging Practice

Charlie has 1,243 stamps. He gives away 12 stamps. His father gives him 415 stamps. He divides as many stamps as possible equally among 4 albums.

Question 1.
How many stamps did he place in each album?

Answer:
He places 411 stamps in each album.

Explanation:
Given that Charlie has 1,243 stamps and he gives away 12 stamps then it will be 1,243 – 12 = 1,231 and his father gives him 415 stamps then it will be 1,231+415 which is 1,646. As he divides as many stamps as possible equally among 4 albums it will be 1,646÷4 which will be approx 411 stamps.

Question 2.
Based on your answer in Exercise 1, how many stamps are left over?
Answer:
2 stamps.

Explanation:
The number of stamps leftover is 2 stamps.

Put on Your Thinking Cap!

Problem Solving

Question 1.
The cost of 2 televisions and 3 DVD players is $1,421. The cost of 1 DVD player is half the cost of 1 television. What is the cost of 1 television?
Answer:

Explanation:
Given that the cost of 2 televisions and 3 DVD players is $1,421 and the cost of 1 DVD player is half the cost of 1 television. Let the cost of 1 DVD player be X, so the price of the television will be 2X. So 2 televisions and 3 DVD player will be 2(2X)+3(X) = $1,421
4X+3X = $1,421
7X = $1,421
X = $203.
So the cost of 1 television it will be 2($203) which is $406.

Go through the Math in Focus Grade 8 Workbook Answer Key Chapter 4 Review Test to finish your assignments.

Math in Focus Grade 8 Course 3 A Chapter 4 Review Test Answer Key

Concepts and Skills

Find the slope of each line using the points indicated. Then write an equation for it.

Question 1.

Answer:
By seeing the above graph we can say that the line passes through (0, 0) and (12,5)
slope = (5 – 0)/(12-0)
slope = 5/12
The line passes through the y-axis at the point (0, 0).
Thus m = 5/12 and y-intercept b is 0.
y = 5/12 x

Question 2.

Answer:
The line passes through the points (4, 0) and (0, 6).
Slope m = \(\frac{6-0}{0-4}\)
= \(\frac{6}{-4}\)
= –\(\frac{3}{2}\)
The line passes through the y-axis at the point (0, 6).
Thus m = –\(\frac{3}{2}\) and y-intercept b is 6.
y = –\(\frac{3}{2}\)x + 6

Question 3.

Answer:
The line passes through the points (2, 3) and (-8, 3).
Slope m = \(\frac{3-3}{-8-2}\)
= \(\frac{0}{-10}\)
= 0
The line passes through the y-axis at the point (0, 3).
Thus m = 0 and y-intercept b is 3.
y = 3

Question 4.

Answer:
The line passes through the points (1, 2) and (1, -2).
Slope m = \(\frac{-2-2}{1-1}\)
undefined

For each line, state its slope and its y-intercept.

Question 5.
y = \(\frac{1}{2}\)x – 3
Answer:
The equation is in the form of y = mx + b
y = \(\frac{1}{2}\)x – 3
m = \(\frac{1}{2}\)
y-intercept = -3

Question 6.
y = -3x + 4
Answer:
The equation is in the form of y = mx + b
y = -3x + 4
m = -3 and
y-intercept = 4

Write an equation of each line given its slope and its y-intercept.

Question 7.
Slope, m = -4
y-intercept, b = –\(\frac{1}{3}\)
Answer:
Given,
Slope, m = -4
y-intercept, b = –\(\frac{1}{3}\)
y = -4x – \(\frac{1}{3}\)

Question 8.
Slope, m = \(\frac{2}{5}\)
y-intercept, b = 3
Answer:
Given,
Slope, m = \(\frac{2}{5}\)
y-intercept, b = 3
y = \(\frac{2}{5}\)x + 3

Solve. Show your work.

Question 9.
Write an equation of the line parallel to 5y = 3x + 12 that has a y-intercept of 2.
Answer:
Given,
5y = 3x + 12
y = \(\frac{3}{5}\)x + 12
y-intercept of 2
y = \(\frac{3}{5}\)x + 2

Question 10.
Write an equation of the line that has slope –\(\frac{1}{2}\) and passes through the point (-4, 5).
Answer:
Given,
slope m = –\(\frac{1}{2}\)
y-intercept = 5
The equation of the line is y = mx + b
y = –\(\frac{1}{2}\)x + 5

Question 11.
Write an equation of the line that passes through the point (-4, -4) and is parallel to 2y – x = -6.
Answer:
Given equation
2y – x = -6
2y = x – 6
y = \(\frac{1}{2}\)x – 3
the line that passes through the point (-4, -4)
y = \(\frac{1}{2}\)x – 4

Question 12.
Write an equation of the line that passes through the point (-4, -3) and is parallel to 4y – x = -16.
Answer:
Given equation
4y – x = -16
4y = x – 16
y = \(\frac{1}{4}\)x – 4
the line that passes through the point (-4, -3)
y = \(\frac{1}{4}\)x – 3

Write an equation of the line that passes through each pair of points.

Question 13.
(0, 0) and (7, 7)
Answer:
The line passes through the points (0, 0) and (7, 7).
Slope m = \(\frac{7-0}{7-0}\)
= \(\frac{7}{7}\)
= 1
The line passes through the y-axis at the point (0, 0).
Thus m = 1 and y-intercept b is 0.
y = x

Question 14.
(1, 2) and (4, 8)
Answer:
The line passes through the points (1, 2) and (4, 8).
Slope m = \(\frac{8-2}{4-1}\)
= \(\frac{6}{3}\)
= 2
Thus m = 2 and y-intercept b is 0.
y = 2x

Use graph paper. Graph each linear equation. Use 1 grid square on both axes to represent 1 unit for the interval from -5 to 5.

Question 15.
4y = -3x – 8
Answer:
Given the equation
4y = -3x – 8
y = –\(\frac{3}{4}\)x – 2
slope = –\(\frac{3}{4}\) and
y-intercept = -2

Question 16.
Slope = \(\frac{1}{3}\); (0, -2)
Answer:
Given,
Slope = \(\frac{1}{3}\); (0, -2)
The equation of the line is y = mx + b
Substitute the given slope and points in the equation.
The equation of the line is y = \(\frac{1}{3}\)x – 2

Problem Solving

Solved Show your work.

Question 17.
Landscaping Company A and Company B each charges a certain amount, C dollars, as consultation fee, plus a fixed hourly charge.

a) Find the amount each landscaping company charges as its consultation fee.
Answer: Company A charges $400 and Company B charges $200.

b) Which company charges a greater amount per hour?
Answer: Company A

Question 18.
The operator of a charter bus service charges a certain amount for a bus, plus per-passenger charge. The graph shows the total charges, C dollars, for carrying x passengers.

a) Find the vertical intercept and explain what information it gives about the situation.
Answer: The vertical intercept is 100

b) Find the slope of the graph and explain what information it gives about the situation.
Answer:
The line passes through the points (40, 400) and (0, 100)
y-intercept is 100.
y = (100 – 400)/(0 – 40)
y = -300/-40
y = 30/4
y = 30/4 x + 100

Go through the Math in Focus Grade 8 Workbook Answer Key Chapter 4 Lesson 4.5 Real-World Problems: Linear Equations to finish your assignments.

Math in Focus Grade 8 Course 3 A Chapter 4 Lesson 4.5 Answer Key Real-World Problems: Linear Equations

Math in Focus Grade 8 Chapter 4 Lesson 4.5 Guided Practice Answer Key

Solve. Show your work.

Question 1.
Jeanette rents a bike while visiting a city. She pays $7 per hour to rent the bike. She also pays $8 to rent a baby seat for the bike. She pays this amount for the baby seat no matter how many hours she rents the bike. The graph shows her total cost, y dollars, after x hours.


a) Find the vertical intercept of the graph and explain what information it gives about the situation.
From the graph, the vertical intercept is . It represents .
Answer:
From the graph, the vertical intercept is 8, It represents the initial cost of $8 to rent a baby seat for the bike.

b) Find the slope of the graph and explain what information it gives about the situation.
The graph passes through (, ) and (, ).
Let (, ) be (x₁, y₁) and (, ) be (x₂, y₂).
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\) Use the slope formula.
= \(\frac{?}{?}\) Substitute values.
= \(\frac{?}{?}\) Subtract.
= Simplify.
The line has slope m = . It represents .
Answer:
From the given graph,
The graph passes through the points as (0, 8) and (8, 64)
Let (0, 8) be x1, y1 and (8, 64) be x2, y2.
So we get the slope of the line as
y2 – y1/ x2 – x1
= 64 – 8/ 8 – 0
= 56/8
= 7
Therefore, the rent charged to a bike is $7/hr.

Solve. Show your work.

Question 2.
Both Zack and Joy are salespeople. Each of them earns a fixed weekly salary and a percent commission based on the total sales he or she makes in a week. The graphs show the total earnings, E dollars, each person can make in one week, based on the person’s total sales, S dollars.

a) Find the fixed weekly salary for each person.
Answer:
From the graph, the vertical intercept is the fixed income of each one.
The monthly income of zack is $600 and Joy is $300.

b) Both Zack and Joy earn a percent commission. Who earns a greater percent in commission?
Answer:
From the graph, Zack earns more commission
Explanation:
From the given graph,
We get the commission of Zack as,
$600 – $300 = $300
Commission of Joy as,
$800 – $600 = $200
So, Zack earns more commission.

c) Find each person’s rate of commission.
Answer:
From the graph,
We get each person’s commission as,
Zack’s commission as $300,
Joy’s commission as $200.

Question 3.
Isaac and George are brothers who live at the same house, but go to different cities for vacation. When their vacation is over, they begin driving back home at the same time, but drive home at different speeds. Isaac’s distance D miles from their house x hours after he starts driving is given by the equation D = -50x + 150. The graph shows George’s distance D miles from their house x hours after he starts driving home.

a) Find the vertical intercept of George’s graph and explain what information it gives about the situation.
Answer:
From the given graph, we get the vertical intercept of George’ graph as 240.
Therefore, George is 240 miles away from home.

b) Find the slope of George’s graph and explain what information it gives about the situation.
Answer:
From the given graph, we get the points as (0,240) and (4,0)
We have the slope of the line as,
y2 – y1/ x2 – x1
= 0 – 240/ 4 – 0
= – 240 / 4
= -60
Here, the slope is negative which means as time increases the distance of George from his house decreases.

c) Which brother is driving faster? How do you know?
Answer:
From the given graph, we get the slope as -60
The equation is given as D = -50x + 150, here the slope is -50,
So the slope for George’s graph is more, therefore he is moving faster.

Math in Focus Course 3A Practice 4.5 Answer Key

Solve. Show your work.

Question 1.
Joe pays a fixed amount each month to use his cell phone. He also pays for each minute that he makes calls on the phone. The graph shows the amount, C dollars, he pays in a given month, based on the airtime, x minutes, he uses to make calls.

a) Find the vertical intercept of the graph and explain what information it gives about the situation.
Answer:
From the given graph,
We have the vertical intercept of the graph as 10,
The fixed deposit for the cell phone is $10.

b) Find the slope of the graph and explain what information it gives about the situation.
Answer:
From the graph, we have the points as (0, 10) and (250, 30)
we get the slope as,
y2 – y1/ x2 – x1
= 30 – 10/ 250 – 0
= 20 / 250
= 2/25
Since slope of the curve is positive which means that when the airtime increases then the charges also increases.

Question 2.
Ricky and Aaron are brothers, and each of them has a coin bank. In January, the boys had different amounts of money in their coin banks. Then, for each month after that, each boy added the same amount of money to his coin bank. The graph shows the amount of savings, S dollars, in each coin bank after t months.

a) Find the initial amount of money in each coin bank.
Answer:
From the graph, we get
The initial amount in Ricky bank account is $30,
The initial amount in Aaron bank account is $10.

b) Who added a greater amount of money each month into his coin bank?
Answer:
From the graph,
For Aaron, the initial saving is $10 and the final saving is $60
So the money added to the bank is $50
For Ricky, the initial saving is $30 and the final saving is $70
So the money added to the bank is $40
Therefore, Aaron added a greater amount of money each month into his coin bank.

Question 3.
Raymond and Randy drive from Town A to Town B in separate cars. The initial amount of gasoline in each car is different. The graphs show the amount of gasoline, y gallons, in each person’s car after x miles.

a) Find the initial amount of gasoline in each car.
Answer:
From the graph,
Initial amount of gasoline in Raymond is 5,
Initial amount of gasoline in Randy is 3.

b) Whose car uses more gasoline?
Answer:
Given, Raymond and Randy drive from Town A and Town B in seperate cars. The graph shows the amount of gasoline, y gallons in each person’s car after x miles.
From the given graph,
Raymond’s line graph passes through the points (0, 5) and (20, 2.5)
Let (0, 5) be x1, y1 and (20, 2.5) be x2 , y2
Slope = y2 – y1 / x2 – x1
= 2.5 – 5 / 20 – 0
= – 2.5/ 20
= – 0.125
From the given graph,
Randy’s line graph passes through the points (0, 3) and (20, 2.5)
Let (0, 3) be x1, y1 and (20, 2.5) be x2, y1
Slope = y2 – y1/x2 – x1
= 2.5 – 3/ 20 – 0
= – 0.5 / 20
= – 0.025
Comparing both values, we get
Raymond’s car uses 0.125 – 0.025 = 0.1 gal of gasoline more than Randy car.

Question 4.
Pete and Winnie visit Star Cafe every day and they pay for the items using a gift card. The amount, y dollars, on Winnie’s gift card after x days is given by the equation y = 100 – 19x. The graph shows the amount on Pete’s gift card over x days.

a) Write an equation for the amount on Pete’s gift card.
Answer:
The equation for the amount on Pete’s gift card is y = 80 – 10x
Explanation:
From the graph given,
Pete’s line graph passes through the points (0, 80) and (5, 30)
Let (0, 80) x1, y1 and (5, 30) x2, y2
Slope = y2 – y1 / x2 – x1
= 30 – 80 / 5 – 0
= – 50/ 5
= – 10
Equation of the line standard form is y – y1 = m(x – x1)
Substituting the values of slope and (x1, y1) = (0, 80) in the equation
y – 80 = – 10(x – 0)
= y – 80 = -10x + 0
= y = 80 – 10x
Hence, the equation for the amount on Pete’s gift card is y = 80 – 10x.

b) Using your answer in a), whose gift card had a higher initial amount?
Answer:
The amount, y dollars on Winnie’s gift card after x days is given by the equation
y = 100 – 19x — (1)
The amount, y dollars on Pete’s gift card after x days is given by the equation
y = 80 – 10x — (2)
Both the equations are in slope-intercept form
As the amount is given by vertical or y-axis, comparing the vertical interceptions of both the equations, we get
100 > 80
Therefore, Winnie’s gift card has a higher initial amount

c) Using your answer in a), who spends more each day?
Answer:
Equation of Winnie’s gift card is given as y = 100 – 9x
We got the equation of Pete’s gift card as y = 80 – 10x
Equation of the line graphs of Winnie and Pete are given by
y = 100 – 19x — (1)
y = 80 – 10x — (2)
Can also be written as
y = -19x + 100 — (3)
y = -10x – 80 — (4)
Putting them in the slope intercept form, i.e., y = mx + b
Slope of lines Winnie’s is -19 and Pete’s is -10
Winnie spends more each day as his slope is greater.

Use graph paper. Solve.

Question 5.
A scientist attaches a spring that is 11 inches long to the ceiling and hangs weights from the spring to see how far it will stretch. The scientist records the length of the spring, y inches, for different weights x pounds.


a) Graph the relationship between the length of the spring for different v weights. Use 1 grid square to represent 1 unit on the horizontal axis for the x interval 0 to 4, and 1 grid square for 2 units on the vertical axis for the y interval 11 to 19.
Answer:
Weight in pounds is represented by the x-axis
Length of spring in inches is represented by the y-axis
Plotting the points in the graph:

b) Find the vertical intercept of the graph and explain what information it gives about the situation.
Answer:
The vertical intercept of the graph is 11,
This is the amount of weight, 11 pounds.
From the graph,
The graph touches the y-axis at (0,11)

c) Find the slope of the graph and explain what information it gives about the situation.
Answer:
The slope of the graph is 2
It represents the rate at which the length of the spring increases for the weight in pounds,
2 inches for each pound of weight.
The graph is given by

It passes through the points (0,11) and (4,19)
Let (0,11) be x1, y1 and (4,19) be x2, y2
Slope = y2 – y1 / x2 – x1
= 19 – 11 / 4 – 0
= 8 / 4
= 2
Therefore, the slope of the graph is 2.

d) Write an equation relating the spring length and the pounds of weights hung from the spring.
Answer:
The equation relating the spring stretch length and the pounds of weights hung from spring is y = 2x + 11
The graph is given by

It passes through the points (0,11) and (4,19)
Let (0,11) be x1, y1 and (4,19) be x2, y2
The slope of the graph is 2,
Substituting the values of slope and x1y1 = (0,11) in the equation, y – y1 = m(x – x1)
y – 11 = 2(x – 0)
y – 11 = 2x
y = 2x + 11
Hence, the required equation of the line is y = 2x + 11.

Brain @ Work

Question 1.
Conrad and Angeline each receive an allowance. Conrad gets the entire week’s allowance on Monday. He spends the same amount every day. Angeline gets a daily allowance starting on Monday. She saves the same amount every day. After four days, both have the same amount of money. The graph shows the amount of money, y dollars, Conrad has after x days during one week.

a) Copy the graph. Then draw a line to represent the amount of money Angeline has after x days.
Answer:

The graph representing the amount of money Angeline has after d days along with that of Conrad is

b) Find the slope of Conrad’s graph and explain what information it gives about the situation.
Answer:
From the above graph,
The line passes through the points (0,28) and (7,0)
Let (0,28) be x1,y1 and (7,0) be x2,y2
Slope = y2 – y1/ x2 – x1
= 0 – 28 / 7 – 0
= -28 / 7
= – 4
The slope of the graph is -4,
It represents the rate at which the amount of money of Conrad keeps on decreasing with each day. Therefore, 4 dollars for each day.

c) Write an equation to represent the amount of money each person has during that week.
Answer:
The equation to represent the amount of money each person has during that week is y = 7 – x
The graph is

The slope of the line of Conrad is -4
The line passes through the points (0,28) and (7,0)
Substituting the values in equation of line,
y – y1 = m(x – x1)
=> y – 28 = -4(x – 0)
=> y = -4x + 28
=> y = -x + 28/4
=> y = -x + 7
=> y = 7 – x

Question 2.
Gordon left Townsville at 12 P.M. and started biking to Kingston 50 miles away. One and a half hours later, Jonathan left Kingston and started biking toward Townsville at a speed of 20 miles per hour. The graph shows Gordon’s distance, d miles, from Townsville after t hours.

a) Copy the graph. Then draw a line to represent Jonathan’s distance from Kingston after t hours.
Answer:
The graph representing Jonathan’s distance from Kingston after t hours is

b) Find the slope of Gordon’s graph and explain what information it gives about this situation.
Answer:
From the above graph,
The line passes through the points (0,0) and (3,50)
Let (0,0) be x1,y1 and (3,50) be x2,y2
Slope = y2 – y1/ x2 – x1
= 50 – 0 / 3 – 0
= 50 / 3
= 16.67
The slope of the graph is 16.67
It represents the rate at which Gordan’s distance increases from Townsville after each hour, i.e., 16.67 miles per hour.

c) Write an equation to represent each person’s distance from Townsville after t hours.
Answer:
The graph is

The slope of the line of Gordon is 16.67
The line passes through the points (0,0) and (3,50)
Substituting the values in the equation of line is
y – y1 = m(x – x1)
=> y – 0 = 16.67(x – 0)
=>y = 16.67x
Therefore, the required equation is y = 16.67x

Go through the Math in Focus Grade 8 Workbook Answer Key Chapter 2 Lesson 2.3 Multiplying and Dividing in Scientific Notation to finish your assignments.

Math in Focus Grade 8 Course 3 A Chapter 2 Lesson 2.3 Answer Key Multiplying and Dividing in Scientific Notation

Math in Focus Grade 8 Chapter 2 Lesson 2.3 Guided Practice Answer Key

Complete.

Question 1.
In the 19th century, the Law Courts of Brussels was the largest building ever built. Its base area measures about 1.6 • 10² meters by 1.5 • 10² meters. Find the approximate base area of the building.
Approximate base area =
Length • Width

The area enclosed by the outer wall is approximately square meters.
Answer:
Given that Brussels was the largest building ever built.
Its base area measures about 1.6 × 102 meters by 1.5 × 102 meters it means
Length = 1.6 × 10²
Width = 1.5 × 10²
We know that
Base area = length × width
Base area = 1.6 × 1 0² × 1.5 × 10²
= 1.6 × 1.5 × 10⁴
= 2.4 × 10⁴ m²
The area enclosed by the outer Wall is approximately 2.4 × 10⁴ square meters.

Question 2.
The outer wall of Angkor Wat, a World Heritage site in Cambodia, encloses an area of about 1.02 • 10³ meters by 8.02 • 10² meters. Find the approximate area enclosed by the outer wall.
Approximate area enclosed by outer wall
= Length • Width

The area enclosed by the outer wall ¡s approximately square meters.
Answer:
Given that the length of a world Heritage site in Cambodia is 1.02 × 10³
The width of a world Heritage site in Cambodia is 8.02 × 10²
Area enclosed by the outer Wall = length × width
= 1.02 × 10³ × 8.02 × 10²
= 1.02 × 8.02 × 10⁵
= 8.1804 × 10⁵
The area enclosed by the outer Wall is approximately 8.1804 × 10⁵

Complete. Round each coefficient answer to the nearest tenth.

Question 3.
The Jean-Luc Lagardere plant in France is the second largest building in the world. It has an approximate volume of 5.6 • 10⁶ cubic meters. The NASA vehicle assembly building in Florida has a volume of about 3.7 • 10⁶ cubic meters. How many times as great as the volume of the NASA vehicle assembly building is the volume of the Jean-Luc Lagardere plant?

The volume of the Jean-Luc Lagardere plant is about times as great as the volume of the NASA vehicle assembly building.
Answer:
Given that the volume of Jean-Luc Lagardere plant in France is 5.6 × 10⁶ cubic meters
And the volume of the NASA vehicle assembly building is 3.7 × 10⁶ cubic meters
Now the volume of Jean-Luc Lagardere plant /volume of NASA vehicles assembly building is 5.6 × 10⁶/3.7 × 10⁶
= 5.6/3.7 × 10⁶/10⁶
= 20.72 cubic meters
The volume of the Jean-Luc Lagardere plant is 20.72 cubic meters great as the volume of the NASA vehicle assembly building.

Question 4.
The Abraj Al-Bait towers in Saudi Arabia has a floor area of about 1.5 • 10⁶ square meters. The Palazzo in Las Vegas has an approximate floor area of 6.5 • 10⁵ square meters. How many times as great as the floor area of the Palazzo is the floor area of the Abraj Al-Bait towers?

The floor area of the Abraj Al-Bait towers is approximately times greater than the floor area of the Palazzo.
Answer:
The floor area of Abraj Al-Bait towers in saudi Arabia is 1.5 × 10⁶ m²
The floor area of Palazzo in Las Vegas is 6.5 × 10⁵ m²
Therefore the floor area of Abraj AL-Bait towers/floor area of Palazzo
= 1.5 × 10⁶/6.5 × 10⁵
= 1.5/ 6.5 × 10⁶/10⁵
= 0.23 × 10¹
= 2.3 m²
The floor area of the Abraj Al-Bait Towers is approximately 2.3 m² times greater than the floor area of the Palazzo.

Math in Focus Course 3A Practice 2.3 Answer Key

Evaluate each expression in scientific notation, and round the coefficient to the nearest tenth.

Question 1.
7.45 • 10⁶ • 5.4 • 10^-6
Answer:
7.45 • 10⁶ • 5.4 • 10-6
Bases are equal powers should be added
= 7.45 • 5.4
= 40.23
40.23 rounded to the nearest tenth is 40

Question 2.
6.84 • 10^-5 • 4.7 • 10¹⁰
Answer:
6.84 • 10-5 • 4.7 • 1010
Bases are equal powers should be added.
6.84 • 4.7 • 10⁵
32.148 • 10⁵
32.148
32.148 rounded to the nearest tenth is 32.1

Question 3.
5.75 • 10^-5 ÷ (7.15 • 10⁷)
Answer:
Given that
5.75 • 10-5 ÷ (7.15 • 107)
Here 10-5 /107 = 1/10²
So, 5.75 ÷ 7.15 × 10-2
0.804 × 10-2
0.804 × 10-2 is rounded to the nearest tenth is 0.8 × 10-2

Question 4.
8.45 • 10¹¹ ÷ (1.69 • 10^-8)
Answer:
Given that
8.45 • 1011 ÷ (1.69 • 10-8)
Here 1011 /10-8 = 10³
So, 8.45 × 10³ ÷ 1.69
5 × 10³
5 × 10³ rounded to the nearest is 5 × 10³

The table shows the approximate volumes of some planets. Use the information to answer questions 5 to 7. Round your answers to the nearest tenth.

Question 5.
About how many times as great as the volume of Mars is the volume of Venus?
Answer:
Given that the volume of Mars is 1.6 × 10¹¹.
And the volume of Venus is 9.4 × 10¹¹
To find which one is great for that subtract the volume of Mars from the volume of Venus
That is 9.4 × 10¹¹/1.6 × 10¹¹
= 9.4/1.6 × 10¹¹/10¹¹
= 5.875
Therefore the volume of Venus is 5.875 times as great as Mars.
5.875 rounded to the nearest tenth of 6

Question 6.
About how many times as great as the volume of Mars is the volume of Earth?
Answer:
Given that the volume of Mars is 1.6 × 10¹¹.
And the volume of earth is 1.1 × 10¹²
To find which one is great for that we subtract the volume of earth from the volume of Mars.
That is 1.6 × 10¹¹ / 1.1 × 10¹²
= 1.6/1.1 × 10¹¹/10¹²
= 1.45 × 1/10¹
= 1.45 × 0.1
= 0.145
Therefore the volume of Mars is 0.145 as great as the Earth.
0.145 rounded to the nearest tenth is 0.1

Question 7.
About how many times as great as the volume of Venus is the volume of Earth?
Answer:
Given that the volume of Venus is 9.4 × 10¹¹
And the volume of earth is 1.1 × 10¹²
To find which one is great for that we subtract the volume of earth from the volume of Mars
That is 9.4 × 10¹¹ / 1.1 × 10¹²
= 9.4/1.1 × 10¹¹/10¹²
= 8.54 × 1/10¹
= 8.54 × 0.1
= 0.854
Therefore the volume of earth is 0.854 as great as Venus.
0.854 rounded to the nearest tenth is 0.9

Solve. Show your work.

Question 8.
Suzanne’s digital camera has a resolution of 2560 • 1920 pixels. Douglas’ digital camera has a resolution of 3264 • 2448 pixels.
a) Express the resolution of the digital cameras in prefix form to the nearest whole unit. Use the most appropriate unit.
Answer:
Suzanne: 2560 ∙ 1920 We are given the resolutions:
Douglas: 3264 ∙ 2448
2560 ∙ 1920 = 4,915, 200 ≈ 4.92 ∙ 10⁶ pixeli a) We express the resolution of Suzannes camera in prefix form:
= 4.92 megapixeli
3264 ∙ 2448 = 7, 990, 272 ≈ 7.99 ∙ 10⁶ pixeli We express the resolution of Douglas camera in prefix form:
= 7.99 megapixeli
4.99 < 7.99 b) the two numbers have the same exponent We compare the coefficients:
4.99. 10⁶ < 7.99. 10⁶

b) Whose camera has a higher resolution?
Answer:
Given that Suzanne’s digital camera has a resolution of 2560 × 1920 pixels.
= 49,15,200
And the Douglas digital camera has a resolution of 3264 × 2448 pixels.
= 79,90,272
Therefore Douglas digital camera has a higher resolution

Question 9.
Bobby downloaded pictures of a cruise ship and a ski run from the internet. The file size of the cruise ship is about 794 kilobytes while the file size of the ski run is about 2.6 megabytes.
a) What is the total file size, in megabytes and in kilobytes, of a file containing the two pictures?
Answer:
Given that the file size of the cruise ship is 794 kilobytes.
And the file size of the ski run is 2.6 megabytes.
The total file size of two pictures is 794 kilobytes + 2.6 megabytes
1 kilobyte = 0.001 megabyte
794 kilobytes = 0.794 megabytes
Therefore 0.794 + 2.6 = 3.394 megabytes.
The total file size of two pictures in megabytes is 3.394 megabytes.
1 megabyte = 1000 kilobytes
2.6 kilobytes = 2.6 × 1000 = 2600 kilobytes
Therefore the total file size of two pictures in kilobytes is 2600 kilobytes.

b) Calculate the difference in file size, in megabytes and in kilobytes, between the two pictures.
Answer:
Given that the file size of the cruise ship is 794 kilobytes.
1 kilobyte = 0.001 megabyte
794 kilobytes = 0.794 megabytes
And the file size of the ski run is 2.6 megabytes.
1 megabyte = 1000 kilobytes
2.6 megabytes = 2.6 × 1000 = 2600 kilobytes
Therefore the file size of the cruise ship is 794 kilobytes.
And the file size of the ski run is 2600 kilobytes.
The difference between the two pictures in kilobytes is 2600 – 794 = 1806
The file size of the cruise ship is 0.794 megabytes
And the file size of the ski run is 2.6 megabytes.
The difference between the two pictures in megabytes is 2.6 – 0.794 = 1.806

c) To the nearest tenth, about how many times as great as the file size of the ski run picture is the file size of the ship picture?
Answer:
Given that the file size of the cruise ship is 794 kilobytes.
Here 794 nearest to the tenth place is 790 kilobytes
1 kilobyte = 0.001 megabyte
790 kilobytes = 0.001 × 790 = 0.79 megabytes
And the file size of the ski run is 2.6 megabytes
2.6 nearest to the tenth is 3 megabytes
Therefore 3 megabytes – 0.79 megabytes
= 2.21 megabytes
The file size of the ski run picture is 2.21 megabytes great as the file size of the cruise ship.

d) Bobby saved the two pictures on a thumb drive with a capacity of 256 megabytes. Find the remaining free capacity of the thumb drive to the nearest tenth megabyte after Bobby saved the two pictures in it.
Answer:
Given that, Bobby saved two pictures on the thumb drive with a capacity of 256 megabytes.
256 nearest to the tenth is
First look at the tens place it is 5 and the digit to the right is 6.
Here the digit 6 is above 5. So, we add 1 to the tens place and place 0 on the ones place.
Therefore 256 nearest to the tenth place is 260

Question 10.
The Georgia Aquarium in Atlanta is about 2.63 • 10³ inches long, 1.26 • 10² inches wide, and 3 • 10¹ inches deep at its largest point. Find its approximate volume.
Answer:
Given that the length of a Georgia Aquarium in Atlanta is 2.63 × 10³.
The width of a Georgia Aquarium in Atlanta is 1.26 × 10².
The depth of the Georgia Aquarium in Atlanta is 3 × 10¹.
We know that the volume formula for a rectangle is length × width × height.
Georgia Aquarium is in the shape of a rectangle
Volume = 2.63 × 10³ × 1.26 × 10² × 3 × 10¹
Bases are equal powers should be added
= 2.63 × 1.26 × 3 × 10⁶
= 9.9414 × 10⁶.

Question 11.
The square base of the Great Pyramid of Khufu has a length of approximately 1.476 • 10³ feet. Its height is about 2.17 • 10² feet. Find the approximate volume of the pyramid. Write your answer in scientific notation. Round the coefficient to the nearest tenth.

Answer:
Given that the length of a Great Pyramid of Khufu is 1.476 × 10³ feet.
And the height of the great Pyramid of Khufu is 2.17 × 10² feet.
Volume of a pyramid = ⅓ × bh
b = 1.476 × 10³
h = 2.17 × 10²
= ⅓ × 1.476 × 10³ × 2.17 × 10²
⅓ × 1.476 × 2.17 × 10⁵
1.067 × 10⁵
= 1,06,700
1,06,700 nearest to the tenth is 106700.

Question 12.
The Tropical Islands Resort is housed inside a former airplane hangar approximately 1.18 • 10³ feet long, 6.89 • 10² feet wide, and 3.51 • 10² feet high. Use the formula for the volume of a rectangular prism to approximate the volume enclosed by the resort. Round the coefficient to the nearest tenth.
Answer:
Given that the length of a tropical island resort is housed inside a former airplane hangar is 1.18 × 10³.
The wide of a tropical island resort housed inside a former airplane hangar is 6.89 × 10².
The height of a tropical island resort housed inside a former airplane hangar is 3.51 × 10².
We know that the volume of the rectangular prism is width × height × length.
Volume of the rectangular prism = 1.18 × 10³ × 6.89 × 10² × 3.51 × 10²
Bases are same powers should be added
1.18 × 6.89 × 3.51 × 10⁷
= 28.537 × 10⁷
= 28.537 × 10⁷ nearest to the tenth is 30 × 10⁷

Question 13.
The time light takes to. travel one meter in a vacuum is about 3.3 nanoseconds. To travel one mile it takes about 5.4 microseconds.
a) Find the difference, in microseconds, between the times taken by light to travel one meter and one mile in a vacuum.
Answer:
Given that, the time light takes to travel one meter in a vacuum is 3.3 nanoseconds.
The time light takes to travel one mile in a vacuum is 5.4 microseconds.
1 microsecond = 1000 nanoseconds
5.4 microseconds = 5.4 × 1000 = 5400
Difference in microseconds is 5400 – 3.3 = 5396.7

b) How many times longer, to the nearest tenth, does it take light to travel one mile than one meter?
Answer:

Question 14.
A spherical particle was found to have a radius of 3.5 • 10^-10 meter.
a) Express the diameter in the prefix form using picometers.
Answer:
Diameter of a spherical D = 2r
D = 2 × 3.5 × 10-10
= 7 × 10-8 × 10-2 meters
0.00000007 millimeters.

b) Use your answer in a), express the circumference in the prefix form using nanometers. Use 3.14 as an approximation for π.
Answer:
We know that the circumference of a spherical is C = 2πr
C = 2 × 3.14 × 3.5 × 10-10
C = 21.98 × 10-1 × 10-9
C = 21.98 × 10-1 nanometers.

Brain @ Work

Question 1.
Find the cube root of 2.7 • 10¹⁰.
Answer:
Given that 2.7 • 1010
The cube root is (2.7 • 10¹⁰) is
= (2.7 • 10,00,00,00,000)
= (27000000000)
= 3000
Therefore the cube root of 2.7 • 1010 is 3000.

Question 2.
Given that a = 3 • 10³ and b = 4 • 10², find each value.
a) 2a + b
Answer:
Given equation is 2a + b
Here a = 3 • 103
b = 4 • 102
Substitute a,b in the above given equation is
2(3 • 103) + (4 • 102)
= 2(3000)+400
= 6000+400
= 6400

b) \(\frac{2a}{b}\)
Answer:
The given equation is 2ab
Here a = 3 • 103
b = 4 • 102
Substitute a,b in the above given equation is
2(3 • 103)×(4 • 102)
2(3000)×(400)
6000×400
2400000

Question 3.
Solve each of the following. Write your answer in scientific notation using the basic unit.
a) 80 micrograms + 200 nanograms
Answer:
Convert micrograms into nanograms
1 microgram equal to 1000 nanograms
So, 80 micrograms = 1000 × 80 = 80,000 nanograms
Therefore 80,000 nanograms + 200 nanograms = 80,200 nanograms

b) 3 gigabytes – 700 megabytes
Answer:
Convert gigabytes into megabytes
1 gigabyte equal to 1024 megabytes
So, 3 gigabytes equal to 3 × 1024 = 3072 megabytes
Therefore, 3072 megabytes – 700 megabytes = 2372 megabytes.

Go through the Math in Focus Grade 8 Workbook Answer Key Chapter 2 Lesson 2.2 Adding and Subtracting in Scientific Notation to finish your assignments.

Math in Focus Grade 8 Course 3 A Chapter 2 Lesson 2.2 Answer Key Adding and Subtracting in Scientific Notation

Math in Focus Grade 8 Chapter 2 Lesson 2.2 Guided Practice Answer Key

Complete.

Question 1.
The population of Washington, D.C., is about 5.9 • 10⁵. South Dakota has a population of approximately 8 • 10⁵.

a) Find the sum of the populations.
Sum of populations
= Population of Washington D.C. + Population of South Dakota

The sum of the populations is about .
Answer:
Sum of populations
= Population of Washington D.C. + Population of South Dakota
5.9 • 10⁵ + 8 • 10⁵.
Substitute the value 10⁵. from each term
Add within the parentheses
(5.9 + 8) • 10⁵.
13.9 • 10⁵.
write 13.9 in the scientific notation.
1.39 • 10¹.• 10⁵.
Use the product of powers property.
1.39 • 10⁶.

b) Find the difference in the populations.
Difference in the populations
= Population of South Dakota – Population of Washington, D.C.

The difference in the populations is about .
Answer:
Population of South Dakota – Population of Washington, D.C.
8 • 10⁵ – 5.9 • 10⁵.
Substitute the value 10⁵. from each term
Add within the parentheses
(8 – 5.9) • 10⁵.
2.1 • 10⁵.
write 13.9 in the scientific notation.
21 • 10¹.• 10⁵.
Use the product of powers property.
21 • 10⁶.

Question 2.
The approximate length of the smallest salamander is 1.7 • 10^-2 meter. The smallest lizard is about 1.6 • 10^-2 meter long.

a) What is the sum of the lengths of the salamander and the lizard?
Sum of the lengths of the salamander and the lizard
= Length of salamander + Length of lizard
= 1.7 • 10^-2 + 1.6 • 10^-2 Substitute.
= ( + ) • Factor from each term.
= • m within parentheses.
The sum of the lengths is about meter.
Answer:
Sum of the lengths of the salamander and the lizard
= Length of salamander + Length of lizard
= 1.7 • 10^-2 + 1.6 • 10^-2 Substitute.
= (1.7 + 1.6) • 10^-2
= (3.3)• 10^-2
The sum of the lengths is about (3.3)• 10^-2 meter.

b) What is the difference ¡n the length of the reptiles?

The salamander is about meter longer than the lizard.
Answer:
Difference in length between the salamander and lizard.
= Length of salamander – Length of lizard
= 1.7 • 10^-2 – 1.6 • 10^-2 Substitute.
= (1.7 – 1.6) • 10^-2
= 0.1 • 10^-2
= 1 • 10^-1 • 10^-2
= 1 • 10^-3

Question 3.
The approximate area of the continent of Australia is 9 • 10⁶ square kilometers.
The area of the continent of Antarctica is about 1,37 • 10⁷ square kilometers.
a) Find the approximate sum of the land areas of the two continents.

The sum of the land areas is about square kilometers.
Answer:
Approximate sum of the land areas of the two continents
Area of Australia + Area of Antarctica
9 • 10⁶ + 1.37 • 10⁷
9 • 10⁶ + 1.37 • 10⁶ • 10¹
(9 + 13.7)  • 10⁶
22.7  • 10⁶
2.27  • 10⁷  square kilometers

b) What is the difference in the areas of the two continents?
Difference in the land areas
= Area of Antarctica – Area of Australia

The land area of Antarctica is about square kilometers larger than the land area of Australia.
Answer:
The difference in the land areas
= Area of Antarctica – Area of Australia
9 • 10⁶ – 1.37 • 10⁷
1.37 • 10⁶ – 9 • 10⁶
(13.7 – 9)  • 10⁶
4.7 • 10⁶ square kilometers

Solve. Write your answers in scientific notation.

Question 4.
A custom-made invitation using a 10-pt card stock is about 2.54 • 10^-4 meter thick. It is placed inside a tissue paper insert that is approximately 6.0 • 10^-6 meter thick.
a) How thick is the invitation when placed in the tissue paper insert?
Answer:
2.54 • 10^-4+ 6.0 • 10^-6 meters thick
2.54 • 10^-4+ 0.06 • 10^-4 meters thick
(2.54 + 0.06) • 10^-4 meters thick
2.6 • 10^-4 meters

b) How much thicker is the invitation than the tissue paper insert?
Answer:
Let x is a variable here representing the number of times thicker.
Thickness of the card stock is x times as thick as tissue.
2.54 • 10^-4= x • 6.0 • 10^-6
x = 2.54 • 10^-4/6.0 • 10^-6
x = 0.423 • 10²
x = 4.23 • 10²

Complete.

Question 5.
On average, Pluto orbits the Sun at a distance of approximately 4,802 gigameters. Uranus’s average distance from the Sun is about 2.992 • 10⁹ kilometers. Which planet is farther from the Sun? How much farther?

Distance of Pluto from the Sun:
Gm = • km Write in scientific notation.
Difference in distance of Pluto and Uranus from the Sun
= Distance of from the Sun – Distance of from the Sun
= • – • Substitute.
= ( – ) • Factor from each term.
= • km within parentheses.
So, is kilometers farther from the Sun.
Answer:
Given,
On average, Pluto orbits the Sun at a distance of approximately 4,802 gigameters.
Uranus’s average distance from the Sun is about 2.992 • 10⁹ kilometers.
1 gigameter = 10⁶ kilometers.
4,802 gigameters = 4,802 • 10⁶ kilometers.
Difference in distance of Pluto and Uranus from the Sun
= Distance of Pluto from the Sun – Distance of Uranus from the Sun
4,802 • 10⁶ kilometers – 2.992 • 10⁹ kilometers
= 4,802 • 10⁶ kilometers – 2,992 • 10⁶ kilometers
= (4802 – 2992) • 10⁶ kilometers
= 1810 • 10⁶
So, Pluto is 1810 • 10⁶ kilometers farther from the Sun.

Math in Focus Course 3A Practice 2.2 Answer Key

Solve. Show your work. Round the coefficient to the nearest tenth.

Question 1.
6.3 • 10^-2 + 4.9 • 10^-2
Answer:
(6.3 + 4.9) • 10^-2
(11.2) • 10^-2
Now, Round the coefficient to the nearest tenth.
11 • 10^-2

Question 2.
7.2 • 10² – 3.5 • 10²
Answer:
7.2 • 10² – 3.5 • 10²
(7.2 – 3.5) • 10²
3.7 • 10²
Now, Round the coefficient to the nearest tenth.
4 • 10²

Question 3.
3.8 • 10³ + 5.2 • 10⁴
Answer:
3.8 • 10³ + 5.2 • 10⁴
3.8 • 10³ + 5.2 • 10³ • 10¹
(3.8 + 52) • 10³
55.8 • 10³
Now, Round the coefficient to the nearest tenth.
56 • 10³

Question 4.
8.1 • 10⁵ – 2.8 • 10⁴
Answer:
8.1 • 10⁵ – 2.8 • 10⁴
81 • 10⁴ – 2.8 • 10⁴
(81 – 2.8) • 10⁴
78.2 • 10⁴
Now, Round the coefficient to the nearest tenth.
78 • 10⁴

Use the table to answer questions 5 to 9.

The table shows the amounts of energy, in Calories, contained in various foods.

Question 5.
Find the total energy in each food combination. Write your answer in scientific notation. Round coefficients to the nearest tenth.
a) Chicken breast and cabbage
Answer:
Chicken breast: 1.71 • 10⁴
cabbage: 2.5 • 10⁴
Total energy in each food combination is
1.71 • 10⁴ + 2.5 • 10⁴
(1.7 + 2.5) • 10⁴
4.2 • 10⁴
4 • 10⁴

b) Cabbage and raw potato
Answer:
cabbage: 2.5 • 10⁴
Raw potato: 7.7 • 10
Total energy in each food combination is
2500 • 10 + 7.7 • 10
2507.7 • 10
2510 • 10

Question 6.
How many more Calories are in chicken breast than in salmon?
Answer:
Chicken breast: 1.71 • 10⁴
Salmon: 1.67 • 10⁵
1.67 • 10⁵ – 1.71 • 10⁴
(16.7 – 1.71) • 10⁴
14.99 • 10⁴

Question 7.
How many more Calories are in salmon than in cabbage?
Answer:
Salmon: 1.67 • 10⁵
cabbage: 2.5 • 10⁴
1.67 • 10⁵ – 2.5 • 10⁴
(16.7 – 2.5) • 10⁴
14.2 • 10⁴

Solve. Show your work.

Question 8.
A flight from Singapore to New York includes a stopover at Hawaii. The distance between Singapore and Hawaii is about 6.7 • 10³ miles. The distance between New York and Hawaii is about 4.9 • 10³ miles. Write each sum or difference in scientific notation.
a) Find the total distance from Singapore to New York.
Answer:
The distance between Singapore and Hawaii is about 6.7 • 10³ miles.
The distance between New York and Hawaii is about 4.9 • 10³ miles.
6.7 • 10³ miles + 4.9 • 10³ miles
(6.7  + 4.9) • 10³ miles
11.6 • 10³ miles

b) Find the difference in the length of the two flights.
Answer:
The distance between Singapore and Hawaii is about 6.7 • 10³ miles.
The distance between New York and Hawaii is about 4.9 • 10³ miles.
6.7 • 10³ miles – 4.9 • 10³ miles
(6.7  – 4.9) • 10³ miles
1.8 • 10³ miles

Question 9.
Angora wool, obtained from rabbits, has fibers with a diameter of about 1 • 10^-6 meter. Cashmere, obtained from goats, has fibers with a diameter of about 1.45 • 10^-5 meter. Write your answers in the appropriate unit in prefix form.
a) Find the sum of the diameters of the two types of fiber.
Answer:
Angora wool, obtained from rabbits, has fibers with a diameter of about 1 • 10^-6 meter.
Cashmere, obtained from goats, has fibers with a diameter of about 1.45 • 10^-5 meter.
1.45 • 10^-5 meter + 1 • 10^-6 meter.
14.5 • 10^-6 meter + 1 • 10^-6 meter.
(14.5 + 1) • 10^-6 meter.
15.5 • 10^-6 meter.
1.55 • 10^-5 meter.

b) How much wider is the cashmere fiber than the angora fiber?
Answer:
Angora wool, obtained from rabbits, has fibers with a diameter of about 1 • 10^-6 meter.
Cashmere, obtained from goats, has fibers with a diameter of about 1.45 • 10^-5 meter.
1.45 • 10^-5 meter – 1 • 10^-6 meter.
14.5 • 10^-6 meter – 1 • 10^-6 meter.
(14.5 – 1) • 10^-6 meter.
13.5 • 10^-6 meter.
1.35 • 10^-5 meter.

The average distances of three planets from the Sun are shown in the diagram. Use this information for questions 10 to 13. Express your answers in kilometers.

Question 10.
What is the closest Mercury comes to Earth when both are at an average distance from the Sun?
Answer:
Mercury: 5.83 • 10¹⁰
Earth: 1.5 • 10⁸ km = 1.5 • 10⁸ • 10³ m = 1.5 • 10¹⁰• 10¹
Difference between the distance of Earth from Sun to distance of Mercury from Sun
15 • 10¹⁰ – 5.83 • 10¹⁰
(15 – 5.83) • 10¹⁰
9.17 • 10¹⁰

Question 11.
What is the closest Saturn comes to Mercury when both are at an average distance from the Sun?
Answer:
Mercury: 5.83 • 10¹⁰
Saturn: 1.43 • 10¹²
Difference between the distance of Saturn from Sun to distance of Mercury from Sun
143 • 10¹⁰ – 5.83 • 10¹⁰
(143 – 5.83) • 10¹⁰
137.17 • 10¹⁰

Question 12.
What is the closest Saturn comes to Earth when both are at an average distance from the Sun?
Answer:
Earth: 1.5 • 10⁸ km = 1.5 • 10⁸ • 10³ m = 1.5 • 10¹¹
Saturn: 1.43 • 10¹²
Difference between the distance of Saturn from Sun to distance of Mercury from Sun
14.3 • 10¹¹ – 1.5 • 10¹¹
(14.3 – 1.5) • 10¹¹
12.8• 10¹¹

Question 13.
Is the distance you found in 12 greater or less than the average distance from Earth to the Sun? Explain.
Answer: Yes the distance of Saturn from the Sun to the distance of Mercury from the Sun is 12.8• 10¹¹

Solve. Show your work.

Question 14.
Factories A and B produce potato chips. They use the same basic ingredients: potatoes, oil, and salt. Last year, each factory used different amounts of these ingredients, as shown in the table.

a) Which factory used more potatoes last year? How many more potatoes did it use?
Answer: By seeing the above table we can say that Factory A has used more potatoes last year than Factor B.
Potato Factory A Used: 4.87 • 10⁶
Potato Factory B Used: 3,309 • 10³
Potato Factory A Used: 4870 • 10³
Potatoes Used Last Year: 4870 • 10³–  3,309 • 10³
(4870 – 3309) • 10³
1561 • 10³

b) Which factory used more oil last year? How much more oil did it use than the other factory?
Answer:
By seeing the above table we can say that Factory B has used more oil last year than Factor A.
Oil Factory A Used: 356,000 = 356 • 10³
Oil Factory B Used: 5.61 • 10⁵ = 561 • 10³
Oil Used Last Year: 561 • 10³–  356• 10³
(561- 356) • 10³
205• 10³

c) Find the total weight of the ingredients used by each factory. Write your answer in scientific notation.
Answer:
Factory A:
Potato Factory A Used: 4870 • 10³
Oil Factory A Used: 356,000 = 356 • 10³
Salt Factory A Used: 2.87  • 10⁵ = 287 • 10³
4870 • 10³ + 356 • 10³ + 287 • 10³
(4870 + 356 + 287) • 10³
5513 • 10³
Factory B:
Potato Factory B Used: 3,309 • 10³
Oil Factory B Used: 561 • 10³
Salt Factory B Used: 193500 = 193.5  • 10³ 
3309 • 10³ + 561 • 10³ + 193.5 • 10³
(3309+ 561 + 193.5) • 10³
4063.5 • 10³

Question 15.
Math journal The approximate population of the following countries in North America in 2011 are shown in the table. Explain how to use scientific notation to find the total population of the countries.

Answer:
Mexico: 110,000,000 = 11 • 10⁷
Haiti: 9,700 000 = 97 • 10⁵
Costa Rica: 4,600,000 = 46 • 10⁵
United States: 310,000,000 = 31 • 10⁷
11 • 10⁷ + 31 • 10⁷ + 97 • 10⁵ + 46 • 10⁵
4 •  10⁷ + 143 • 10⁵